What is the quadratic formula?

For \(ax^{2} + bx + c = 0\) with \(a \neq 0\), the solutions are \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Both the \(+\) and the \(-\) give a solution, so a quadratic has up to two.

What is the discriminant and what does it tell me?

The discriminant is \(b^{2} - 4ac\), the part under the radical. If it is positive, there are two distinct real solutions; if it is zero, one repeated real solution; if it is negative, two complex-conjugate solutions. It also tells you whether the quadratic factors over the rationals — it does exactly when the discriminant is a perfect square.

Do I need to put the equation in standard form first?

Yes. The formula reads \(a\), \(b\), and \(c\) off the standard form \(ax^{2} + bx + c = 0\). Move every term to one side so the other side is zero before identifying the coefficients.

How do I handle the signs of a, b, and c?

Include the sign as part of the coefficient. In \(2x^{2} - 5x - 3 = 0\), \(a = 2\), \(b = -5\), \(c = -3\). Then \(-b = 5\) and \(b^{2} = 25\). Most quadratic-formula errors are dropped negative signs at this step.

When should I use the quadratic formula versus factoring?

Factor first if the factors are obvious. If you cannot find them within a few seconds, use the formula — it is not much slower and it never fails. The formula is also the safe default under time pressure on a test.

How do I simplify an answer like x = (4 ± √20) / 2?

Simplify the radical first: \(\sqrt{20} = 2\sqrt{5}\), giving \(\dfrac{4 \pm 2\sqrt{5}}{2}\). Then divide every term by the common factor: \(2 \pm \sqrt{5}\). Reduce only when a factor divides the whole numerator.

What if the equation has no b term, like 3x² − 12 = 0?

The formula still works with \(b = 0\): the \(-b\) and \(b^{2}\) simply vanish. But it is usually faster to use the square-root property — \(3x^{2} = 12\), \(x^{2} = 4\), \(x = \pm 2\).

Can I use the quadratic formula when the equation factors?

Yes, always. It returns the same roots, just with more arithmetic. If you can spot the factors in a few seconds, factoring is quicker; if not, the formula is the reliable default and never fails.

What is the quadratic formula?

For \(ax^{2} + bx + c = 0\) with \(a \neq 0\), the solutions are \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Both the \(+\) and the \(-\) give a solution, so a quadratic has up to two.

What is the discriminant and what does it tell me?

The discriminant is \(b^{2} - 4ac\), the part under the radical. If it is positive, there are two distinct real solutions; if it is zero, one repeated real solution; if it is negative, two complex-conjugate solutions. It also tells you whether the quadratic factors over the rationals — it does exactly when the discriminant is a perfect square.

Do I need to put the equation in standard form first?

Yes. The formula reads \(a\), \(b\), and \(c\) off the standard form \(ax^{2} + bx + c = 0\). Move every term to one side so the other side is zero before identifying the coefficients.

How do I handle the signs of a, b, and c?

Include the sign as part of the coefficient. In \(2x^{2} - 5x - 3 = 0\), \(a = 2\), \(b = -5\), \(c = -3\). Then \(-b = 5\) and \(b^{2} = 25\). Most quadratic-formula errors are dropped negative signs at this step.

When should I use the quadratic formula versus factoring?

Factor first if the factors are obvious. If you cannot find them within a few seconds, use the formula — it is not much slower and it never fails. The formula is also the safe default under time pressure on a test.

How do I simplify an answer like x = (4 ± √20) / 2?

Simplify the radical first: \(\sqrt{20} = 2\sqrt{5}\), giving \(\dfrac{4 \pm 2\sqrt{5}}{2}\). Then divide every term by the common factor: \(2 \pm \sqrt{5}\). Reduce only when a factor divides the whole numerator.

What if the equation has no b term, like 3x² − 12 = 0?

The formula still works with \(b = 0\): the \(-b\) and \(b^{2}\) simply vanish. But it is usually faster to use the square-root property — \(3x^{2} = 12\), \(x^{2} = 4\), \(x = \pm 2\).

Can I use the quadratic formula when the equation factors?

Yes, always. It returns the same roots, just with more arithmetic. If you can spot the factors in a few seconds, factoring is quicker; if not, the formula is the reliable default and never fails.

The Quadratic Formula

Factoring is fast when it works, and completing the square shows why the method works — but for actually solving an arbitrary quadratic, the quadratic formula is the workhorse. It is the result of completing the square on the general equation , so it produces the exact solutions of any quadratic, factorable or not, real or complex.

This lesson is about using it without error: getting to standard form, reading , , with their signs, simplifying the radical answer, and — before any of that — reading the discriminant to know what kind of answer to expect.

The Quadratic Formula

Here , , and come from the standard form . The produces the two solutions. The expression under the radical, , is the discriminant.

The Discriminant: Predicting the Solutions

Compute first and you know the outcome before solving:

Discriminant	Solutions
, perfect square	two distinct rational solutions (the quadratic factors)
, not a perfect square	two distinct irrational solutions
	one repeated real solution
	two complex-conjugate solutions

Using the Formula: The Routine

Standard form. Move every term to one side; the other side is .
Identify , , with signs included.
Compute the discriminant .
Substitute into .
Simplify the radical, then reduce any common factor across the whole fraction.

Reading a, b, and c without sign errors

The coefficient carries its sign with it. In , it is , , — not . Then, working through the formula:

(a squared quantity is never negative)
(the two minus signs multiply to a plus)

Almost every wrong answer from the quadratic formula is a dropped negative in one of these three places. Write , , down explicitly, with their signs, before touching the formula.

Simplifying the radical answer

Two things can usually be tidied in :

Simplify the square root. may have a perfect-square factor: ; .
Reduce the whole fraction — but only if one number divides every term of the numerator. reduces to because divides and . does not reduce, because does not divide .

When there is no b term or no c term

The formula still applies, but there is usually a faster route. With no term (), use the square-root property: . With no constant (), factor out : or .

Worked Example A: Two Rational Solutions

Solve .

Coefficients: , , .

Discriminant: . It is a perfect square, so expect two rational answers.

Substitute:

Answer: or .

Worked Example B: Two Irrational Solutions

Solve .

Coefficients: , , .

Discriminant: . Positive, not a perfect square — two irrational answers.

Substitute:

Simplify: , so

Answer: .

Worked Example C: One Repeated Solution

Solve .

Coefficients: , , .

Discriminant: . Exactly one solution.

Substitute:

Answer: , a double root. (Indeed .)

Worked Example D: Complex Solutions

Solve .

Coefficients: , , .

Discriminant: . Negative — two complex solutions.

Substitute:

Answer: .

Worked Example E: Rearranging Before Applying the Formula

Solve .

Step 1 — move every term to one side so the other side is zero:

Coefficients: , , .

Discriminant: . Positive, not a perfect square — two irrational solutions.

Substitute:

Simplify: , and divides , , and :

Answer: . Reading as instead of , or forgetting to move the over first, are the two ways this problem goes wrong.

Common Mistakes to Avoid

Not writing standard form first. must become before reading , .
Dropping a negative sign. If , then and . Track the sign carefully.
Only dividing part of the numerator by . The entire is over .
Simplifying incorrectly. reduces to only because divides every term of the numerator. does not reduce.
Forgetting includes the sign of . With , .
Using the formula on a non-quadratic. must be nonzero; if , the equation is linear.

Where This Shows Up Later

Quadratic Applications. Word problems that end in a messy quadratic are finished with the formula.
Polynomial and Rational Inequalities. Finding where a quadratic factor changes sign starts with its roots from the formula.
Graphing. The roots are the parabola’s -intercepts; the discriminant tells you whether the parabola crosses the -axis, touches it, or misses it.
Equations Reducible to Quadratic Form. After the substitution , the formula solves the resulting quadratic in .
Complex numbers. A negative discriminant is the standard entry point to complex solutions.

Practice Problems

Work each problem yourself before opening the answer.

Problem 1. Solve .

Show answer

, , ; discriminant .

Problem 2. Solve .

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, , ; discriminant .

Problem 3. Solve .

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Discriminant .

Problem 4. Solve .

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Discriminant .

Problem 5. Solve .

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Discriminant .

Problem 6. Solve .

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Discriminant .

Problem 7. Solve .

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Standard form: ; discriminant .

Problem 8. Solve .

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Discriminant .

Problem 9. Solve .

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Discriminant .

Problem 10. Solve .

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Discriminant .

Problem 11. Without solving, state the number and type of solutions of .

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Discriminant : two complex-conjugate solutions.

Problem 12. Solve .

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Discriminant .

Problem 13. Solve .

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Standard form: ; , , . Discriminant .

Problem 14. Solve .

Show answer

Discriminant .

Quick Reference

Item	Value
Formula
Discriminant
, perfect square	two rational solutions (factors)
, not perfect square	two irrational solutions
	one repeated real solution
	two complex solutions
Before using	put the equation in standard form; keep signs on

The formula is Completing the Square solved once for all; the other methods and the method-choosing guide are in Quadratic Equations. Word problems that land here are in Quadratic Applications. More Algebra lessons are available too.

The Quadratic Formula

The Quadratic Formula

The Discriminant: Predicting the Solutions

Using the Formula: The Routine

Reading a, b, and c without sign errors

Simplifying the radical answer

When there is no b term or no c term

Worked Example A: Two Rational Solutions

Worked Example B: Two Irrational Solutions

Worked Example C: One Repeated Solution

Worked Example D: Complex Solutions

Worked Example E: Rearranging Before Applying the Formula

Common Mistakes to Avoid

Where This Shows Up Later

Practice Problems

Quick Reference

Frequently Asked Questions

Related lessons