Quadratic equation solver with steps

The part under the square root, $b^2 - 4ac$, is the discriminant:

• $a = 2$, $b = -8$, $c = 6$.
• Discriminant: $(-8)^2 - 4(2)(6) = 64 - 48 = 16$, positive, so two solutions.
• $x = \dfrac{8 \pm \sqrt{16}}{4} = \dfrac{8 \pm 4}{4}$, so $x = 1$ or $x = 3$.
• Factorised: $2(x - 1)(x - 3) = 0$.

A quicker route here: divide by 2 first to get $x^2 - 4x + 3 = 0$, then factorise. Spotting a common factor is always worth a second.

When the discriminant isn't a perfect square, the answers are irrational. For $x^2 - 2x - 1 = 0$, the discriminant is 8, so $x = \dfrac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$. Many exam questions ask for answers "in surd form" or "to 2 decimal places"; the solver gives both.

• Factorising: fastest when the roots are whole numbers or simple fractions.
• Quadratic formula: always works.
• Completing the square: gives the turning point directly and is needed for some proofs.

See the quadratic equations topic guide for practice, and SAT Math formulas for the formulas the SAT expects you to know.

The solutions of $ax^2 + bx + c = 0$ are where the graph of $y = ax^2 + bx + c$ crosses the x-axis. The turning point is halfway between them, at $x = -\tfrac{b}{2a}$. If $a$ is positive the parabola opens upwards (a minimum); if negative, downwards (a maximum). Sketching the graph is a quick way to check your algebra: two crossings, one touch or none should match the discriminant.