Standard form
Awaiting solutionPaste an equation or enter a, b, and c for ax² + bx + c = 0 to get exact roots, rounded decimals, the discriminant, a graph, and full working.
Accepts fractions, decimals, √ radicals, parentheses, implicit multiplication, and terms on either side. Use sqrt(2) or √2. Without =, the expression is set equal to zero.
Keyboard: arrow keys change input tabs; Ctrl/Cmd + Enter solves.
Steps will appear here.
Awaiting solutionAwaiting solutionAwaiting solution
A quadratic equation is any equation that can be written in the form
ax² + bx + c = 0 where a, b, and c are real numbers and
a ≠ 0. Quadratics appear everywhere: projectile motion, optimization
problems, area/geometry questions, and many modeling tasks in science and engineering.
Every quadratic can be solved using the quadratic formula:
x = [-b ± √(b2 - 4ac)] / (2a)
The expression under the square root, b² − 4ac, is called the
discriminant (Δ). It determines the nature of the roots:
a, b, and c.Everything runs client-side in your browser, so your inputs never leave your device.
First move every term to one side. In x² + 4 = 5x, standard form is x² − 5x + 4 = 0, so a=1, b=−5, c=4. A missing term has coefficient zero: 2x²−7=0 has b=0.
Factor when simple factor pairs are visible. Complete the square to expose vertex form or derive the formula. Use the quadratic formula when factors are unclear, or roots are irrational or complex.
Δ > 0 gives two x-intercepts, Δ = 0 gives one tangent intercept, and Δ < 0 gives none. The vertex is at x=−b/(2a); the sign of a determines whether the parabola opens up or down.
Substitute a root into both sides of the original equation. You can also check both roots with Vieta: r₁+r₂=−b/a and r₁r₂=c/a.
x²−5x+6=0: Δ = 25−24 = 1. Thus x=(5±1)/2, so x=3 or x=2.
Vieta check: 3+2=5=−b/a and 3·2=6=c/a.
x²−6x+9=0: Δ = 36−36 = 0. Thus x=6/2=3, repeated.
Substitution: 3²−6·3+9=9−18+9=0.
2x²−8x+5=0: Δ = 64−40 = 24. Then x=(8±√24)/4=(4±√6)/2.
Vieta check: the sum is 4=−b/a; the product is (16−6)/4=5/2=c/a.
x²+4x+8=0: Δ = 16−32 = −16. Then x=(−4±4i)/2=−2±2i.
Vieta check: the sum is −4; the product (−2+2i)(−2−2i)=4+4=8.
(3/4)x²−(5/2)x+2=0. Multiply by 4: 3x²−10x+8=0. Δ = 100−96 = 4, so x=(10±2)/6: x=2 or x=4/3.
Vieta check: 2+4/3=10/3=−b/a and 2·4/3=8/3=c/a.
You may enter terms on either side, such as x²+4=5x. The solver subtracts the right side and shows the resulting standard form.
For x²−9=0, enter a=1, b=0, c=−9 in coefficient mode—or simply omit the x term in equation mode.
(4+√6)/2 is exact. Its decimal is rounded, so use the exact form in symbolic work and the decimal for measurement or plotting.
Real roots are x-intercepts because they make the polynomial’s y-value zero. Complex roots do not appear as intercepts on a real-coordinate graph.
If a=0, the solver identifies the input as linear and solves bx+c=0 instead of applying a quadratic formula that would divide by zero.
x = [−b ± √(b² − 4ac)]/(2a). It solves ax²+bx+c=0 for x when a ≠ 0.
Δ=b²−4ac. If Δ>0 there are two real roots; if Δ=0 there is one repeated real root; if Δ<0 there are two complex roots.
Yes. Use forms such as 3/4, -0.5, or (1/2)x². Finite decimals are kept as exact fractions for algebra; radicals in coefficients are accepted and evaluated.
The expression is not quadratic. The calculator solves the remaining linear equation bx+c=0, or reports an identity or contradiction when b is also zero.
Omit the term in equation mode, or enter 0 for its coefficient. For x²+5=0, b=0; for x²−3x=0, c=0.
When b²−4ac is negative, the graph does not cross the x-axis and the two solutions are complex conjugates.
An exact answer preserves fractions and simplified radicals. A decimal answer is a rounded approximation, shown to up to 10 decimal places.
Factoring is usually fastest when integer or simple rational factors are visible. The quadratic formula always works and is preferable for irrational or complex roots.
Substitute each root into the original equation, or check Vieta’s formulas: the sum is −b/a and the product is c/a.
A root makes y=ax²+bx+c equal zero. On the graph, y=0 is the x-axis, so real roots occur exactly where the parabola meets that axis.