Free video lesson

How to Use the Quadratic Formula on the SAT

The quadratic formula solves any quadratic equation — no factoring required. On the SAT, knowing when to use it and how to set it up quickly saves time and prevents errors.

Math · Nonlinear equations · one variable3:02Published July 6, 2026

On YouTube: SAT Quadratic Formula: One Method That Works Every Time

What this lesson covers

✅ What you'll learn:

  • How to identify when the quadratic formula applies
  • How to set up the formula quickly under test conditions
  • Common SAT variations and how to handle them

Worked examples

The questions the video works, written out: the setup, each step, the answer and the trap.

0:32Example 1

Solve x2−6x+7=0x^2 - 6x + 7 = 0.

  1. Try factoring first. You need two integers that multiply to 7 and add to -6. Since 7 is prime, the only pairs are 1 and 7 or -1 and -7, which add to 8 or -8. It does not factor, so use the quadratic formula: x=−b±b2−4ac2a\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  2. Read off a = 1, b = -6, c = 7. Then -b = 6, and under the root b2−4ac=(−6)2−4(1)(7)=36−28=8b^2 - 4ac = (-6)^2 - 4(1)(7) = 36 - 28 = 8.
  3. x=6±82\displaystyle x = \frac{6 \pm \sqrt{8}}{2}. Simplify the root: 8=4⋅2=22\displaystyle \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}, so x=6±222=3±2\displaystyle x = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}.
  4. Check: the two roots must add to -b/a = 6 and multiply to c/a = 7. Sum: (3+2)+(3−2)=6\displaystyle (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6. Product: (3+2)(3−2)=9−2=7\displaystyle (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7.

Answer: x=3±2\displaystyle x = 3 \pm \sqrt{2}

The formula works on any quadratic, including the ones that refuse to factor. The first trap is squaring b: (−6)2(-6)^2 is +36, not -36, and -36 makes the discriminant negative and the problem look unsolvable. The second is dividing only the root by 2a; the 2 underneath divides the 6 as well, which is how 6 becomes 3.

1:01Example 2

How many real solutions does x2−6x+c=0x^2 - 6x + c = 0 have when c = 7, when c = 9, and when c = 11?

  1. Only the discriminant matters: D=b2−4acD = b^2 - 4ac. Positive means two real solutions (the parabola crosses the x-axis twice), zero means exactly one (it touches), negative means none (it misses).
  2. c = 7: D=36−28=8D = 36 - 28 = 8. Positive, so two solutions.
  3. c = 9: D=36−36=0D = 36 - 36 = 0. Exactly one solution. In fact x2−6x+9=(x−3)2x^2 - 6x + 9 = (x - 3)^2, a double root at x = 3.
  4. c = 11: D=36−44=−8D = 36 - 44 = -8. Negative, so no real solutions.
  5. Check with the vertex: the vertex sits at x = 3, where y = 9 - 18 + c = c - 9. That is -2, 0, and 2 for the three values of c, so the graph dips below the axis, then touches it, then floats above it.

Answer: two, one, and zero real solutions

Raising c slides the parabola straight up without moving it sideways, and the discriminant tracks that motion, shrinking from 8 to 0 to -8. The trap is reading D = 0 as zero solutions; a zero discriminant means the plus-or-minus adds nothing, so there is exactly one solution, x = 3.

1:48Example 3

How many real solutions does 2x2+4x−3=02x^2 + 4x - 3 = 0 have, and what are they?

  1. a = 2, b = 4, c = -3. Discriminant: D=42−4(2)(−3)=16+24=40D = 4^2 - 4(2)(-3) = 16 + 24 = 40. Watch the sign: minus four times a negative c becomes plus 24.
  2. D is positive, so there are two real solutions before any solving.
  3. x=−4±402(2)=−4±404\displaystyle x = \frac{-4 \pm \sqrt{40}}{2(2)} = \frac{-4 \pm \sqrt{40}}{4}. Simplify: 40=210\displaystyle \sqrt{40} = 2\sqrt{10}, so x=−4±2104=−2±102\displaystyle x = \frac{-4 \pm 2\sqrt{10}}{4} = \frac{-2 \pm \sqrt{10}}{2}.
  4. As decimals, 10≈3.162\displaystyle \sqrt{10} \approx 3.162, so x is about 0.58 or about -2.58.
  5. Check: the roots must add to -b/a = -2 and multiply to c/a = -3/2. Sum: −2+102+−2−102=−2\displaystyle \frac{-2 + \sqrt{10}}{2} + \frac{-2 - \sqrt{10}}{2} = -2. Product: (−2)2−104=−64=−32\displaystyle \frac{(-2)^2 - 10}{4} = \frac{-6}{4} = -\frac{3}{2}.

Answer: two solutions, x=−2±102\displaystyle x = \frac{-2 \pm \sqrt{10}}{2}, about 0.58 and -2.58

Because c is negative, the -4ac term is positive and the discriminant grows to 40, so two solutions is certain before the formula is even finished. The trap is computing 16 - 24 = -8 and declaring no solutions. Also, since a = 2, the denominator is 4, not 2; dividing by 2 gives answers twice as large as they should be.

Lesson transcript

The narration of the video, word for word, under its chapter headings.

Welcome to SAT Climb. Some quadratics refuse to factor — and those are the ones the SAT uses to burn your clock. One formula cracks every quadratic, and one piece of it tells you how many answers exist before you even solve. Take x² − 6x + 7. Try to factor it: you need two whole numbers that multiply to 7 and add to −6. Seven is prime, so the only pair is 1 and 7 — and they add to 8. No clean factors. Time for the tool that never fails. The quadratic formula. For ax² + bx + c, x = (−b ± √(b²−4ac)) ⁄ 2a. Here a = 1, b = −6, c = 7. Negative b is 6. Under the root, 36 − 28 = 8. So x = (6 ± √8) ⁄ 2, which simplifies to 3 ± √2. Now the shortcut. That piece under the root, b²−4ac, is the discriminant. Its sign alone tells you the number of real solutions. Positive means two: the parabola crosses the x-axis twice. Zero means exactly one: it just touches. Negative means none: it floats clear of the axis. Keep b at −6 and slide c up: 7 gives discriminant 8, two solutions; 9 gives 0, one solution; 11 gives −8, none. Your turn. 2x² + 4x − 3. Start with the discriminant. 16 + 24 = 40 — positive, so two solutions. You'll land on (−2 ± √10) ⁄ 2. Three traps to dodge. One: b² is always positive, so (−6)² is 36, not −36. Two: the 2a underneath divides everything above it, not just the root. Three: a discriminant of zero means one solution, not zero. Clear those, and every quadratic is yours. The quadratic formula: solved. The discriminant reads the future — two, one, or none — before you lift your pencil. Start your free trial at satclimb.com. Keep climbing.

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