How to Find the Discriminant of a Quadratic Equation (Step by Step)

The discriminant of a quadratic equation ax² + bx + c = 0 is the expression b² − 4ac — the piece that lives under the square root in the quadratic formula. To find it, identify a, b, and c, then compute b² − 4ac. Its sign tells you how many real solutions the equation has.

If the discriminant is positive, there are two real solutions. If it is zero, there is exactly one (a repeated root). If it is negative, there are no real solutions — the two solutions are complex.

Worked example: discriminant of 2x² − 5x − 3 = 0

Start by lining up the coefficients with the standard form ax² + bx + c = 0.

• a = 2
• b = −5
• c = −3

Now substitute into b² − 4ac:

1. b² = (−5)² = 25
2. 4ac = 4 · 2 · (−3) = −24
3. b² − 4ac = 25 − (−24) = 25 + 24 = 49

The discriminant is 49. Because 49 > 0 and 49 is a perfect square, the equation has two real, rational solutions (in fact, x = 3 and x = −1/2).

The discriminant formula, step by step

1. Write the equation in standard form. Move every term to one side so it reads ax² + bx + c = 0. Example: x² = 4x − 1 becomes x² − 4x + 1 = 0.
2. Identify a, b, and c with their signs. In x² − 4x + 1 = 0, a = 1, b = −4, c = 1. The sign in front of a term belongs to the coefficient.
3. Square b. Squaring always gives a non-negative result. For b = −4, b² = 16.
4. Compute 4ac. Multiply carefully, watching signs. Here 4 · 1 · 1 = 4.
5. Subtract: b² − 4ac. 16 − 4 = 12. That is the discriminant.

What the value of b² − 4ac tells you

• b² − 4ac > 0 → two distinct real solutions (the parabola crosses the x-axis twice).
• b² − 4ac = 0 → one repeated real solution (the parabola just touches the x-axis).
• b² − 4ac < 0 → no real solutions (the parabola never touches the x-axis).

Second example: discriminant of 3x² + x − 5 = 0

Read off the coefficients:

• a = 3
• b = 1
• c = −5

Substitute into b² − 4ac:

1. b² = 1² = 1
2. 4ac = 4 · 3 · (−5) = −60
3. b² − 4ac = 1 − (−60) = 61

The discriminant is 61. Since 61 > 0 but is not a perfect square, the equation has two real but irrational solutions — you would finish with the quadratic formula to get x = (−1 ± √61) / 6.

Common mistakes when computing b² − 4ac

• Dropping the sign of b. If b = −5, then b² = 25, not −25. Squaring kills the sign.
• Forgetting to move everything to one side first. You cannot read a, b, c off 2x² − 5x = 3 directly — rewrite it as 2x² − 5x − 3 = 0 so that c = −3.
• Sign errors in 4ac. When a or c is negative, 4ac is negative, and subtracting a negative adds. This is where most students lose the answer.
• Confusing the discriminant with the solutions. b² − 4ac is a number, not an x-value. It tells you about the solutions; it is not one of them.
• Assuming a = 1 by default. Always check the coefficient of x². In 2x² − 5x − 3, a is 2, not 1.

Now that you have seen the method on two examples, the fastest way to lock it in is to try one yourself with a guide that catches slips in real time. Sign up for LernOS to work through discriminant problems with a step-by-step tutor that asks the right question at the right moment — instead of just handing you the answer.