Quadratic Formula Calculator – Solve ax² + bx + c = 0
Solve any quadratic equation ax² + bx + c = 0 instantly. Enter the coefficients a, b and c, and the calculator finds both real or complex roots using the quadratic formula and discriminant.
Solve Quadratic Equations
Quadratic Equation Formula
Where:
- \(a\) — coefficient of \(x^2\) (must not be zero)
- \(b\) — coefficient of \(x\)
- \(c\) — constant term
- \(D = b^2 - 4ac\) — the discriminant, which determines the type of the roots
The formula gives the two solutions of the quadratic equation \(ax^2 + bx + c = 0\). The discriminant decides what kind of roots you get:
- \(D > 0\) — two distinct real roots
- \(D = 0\) — one real double root, \(x_1 = x_2 = -\frac{b}{2a}\)
- \(D < 0\) — two complex conjugate roots \(\frac{-b}{2a} \pm \frac{\sqrt{-D}}{2a}i\)
For a deeper treatment of this term under the square root, see the discriminant of a quadratic equation.
Quadratic Equation – Calculation Example
Example: solve \(x^2 - 3x + 2 = 0\), so \(a = 1\), \(b = -3\), \(c = 2\).
\[ D = (-3)^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1 \]
\[ x_{1,2} = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 1} = \frac{3 \pm 1}{2} \]
\[ x_1 = \frac{3 + 1}{2} = 2, \quad x_2 = \frac{3 - 1}{2} = 1 \]
Result: \(x_1 = 2.0000\), \(x_2 = 1.0000\). Check: \(2^2 - 3 \cdot 2 + 2 = 0\) and \(1 - 3 + 2 = 0\).
How to Use the Calculator
- Enter the coefficient of \(x^2\) in the field a (coefficient of x²). It must not be zero.
- Enter the coefficient of \(x\) in the field b (coefficient of x).
- Enter the constant term in the field c (constant).
- Read the results in the fields x₁ and x₂. Real roots are shown as decimal numbers; complex roots appear in the form \(p + qi\) and \(p - qi\).
What Is a Quadratic Equation?
A quadratic equation is any equation that can be written in the standard form \(ax^2 + bx + c = 0\), where \(a \neq 0\). The graph of the corresponding function \(y = ax^2 + bx + c\) is a parabola: it opens upward when \(a > 0\) and downward when \(a < 0\). The real roots of the equation are exactly the points where this parabola crosses the x-axis, which is why solving quadratics is the standard way to find intersections, maximum or minimum values, and optimal points in applied problems.
Quadratic equations appear throughout algebra, geometry, physics, engineering and economics — projectile motion, area and dimension problems, revenue optimization and circuit analysis all reduce to finding roots of a second-degree polynomial. The quadratic formula is guaranteed to work for every quadratic equation, even when factoring is difficult or impossible, which makes it the most reliable solving method.
The nature of the answer is decided by the discriminant \(D = b^2 - 4ac\). If the calculator returns two different numbers, the parabola crosses the x-axis twice. If both results are equal, the parabola just touches the axis at its vertex. If the results contain \(i\), there are no real solutions: the parabola never reaches the x-axis, and the two roots are complex conjugates of each other. You can explore such values further with the complex number calculator.
Assumptions, Limitations and Tips
- The coefficient \(a\) must be non-zero. If \(a = 0\), the equation is linear, not quadratic, and the calculator reports "Not a quadratic equation".
- Results are rounded to 4 decimal places. For most practical purposes this is more than enough, but keep rounding in mind if you substitute the roots back into the original equation for verification.
- If your equation is not in standard form, rearrange it first: move every term to the left side so that \(ax^2 + bx + c = 0\), and only then read off the coefficients. Watch the signs — entering \(-3\) instead of \(3\) for \(b\) changes the answer completely.
- A quick way to check the result: by Vieta's formulas, \(x_1 + x_2 = -\frac{b}{a}\) and \(x_1 x_2 = \frac{c}{a}\). If both hold, the roots are correct.
- For higher-degree polynomials, try the cubic equation solver or the quartic equation solver. If you want to expand a factored form back into standard form first, the FOIL method calculator is useful.
Frequently Asked Questions
How do I solve a quadratic equation by hand without the formula?
Two common alternatives exist. Factoring: look for two numbers that multiply to \(ac\) and add to \(b\); for example \(x^2 - 5x + 6\) factors as \((x-2)(x-3)\), giving roots 2 and 3. Completing the square: rewrite the equation as \(a(x + \tfrac{b}{2a})^2 = \tfrac{b^2-4ac}{4a}\), then take the square root of both sides. Both methods give the same roots — the quadratic formula is just completing the square done once for every equation.
What do complex roots mean graphically and in real problems?
Complex roots mean the parabola never touches the x-axis: it lies entirely above it (if \(a > 0\)) or below it (if \(a < 0\)). In practical problems like projectile motion, a complex root signals that the described event is impossible — for example, a ball thrown with given speed never reaches a certain height. The roots \(p \pm qi\) still satisfy the equation algebraically, which is useful in engineering and signal analysis.
What is a double (repeated) root and why does it matter?
A double root occurs when the discriminant equals zero, so \(x_1 = x_2 = -\tfrac{b}{2a}\). Geometrically, the parabola touches the x-axis at exactly one point — its vertex — without crossing it. This is a boundary case: for example, it tells you the exact height a projectile just barely reaches, or the price at which revenue is exactly at its maximum. Changing \(c\) slightly in either direction splits the root into two real roots or turns both complex.
How do I convert an equation that is not in standard form?
Move every term to the left side so the equation reads \(ax^2 + bx + c = 0\), then combine like terms. For example, \(2x^2 = 5 - 3x\) becomes \(2x^2 + 3x - 5 = 0\), so \(a = 2\), \(b = 3\), \(c = -5\). If you have fractions, multiply through by the common denominator first; if you have parentheses, expand them before reading off the coefficients. The right-hand side must be zero before you enter anything.
How can I find the vertex of the parabola from the coefficients?
The vertex lies at \(x = -\tfrac{b}{2a}\), which is always halfway between the two roots (real or complex). Substitute this value back into \(y = ax^2 + bx + c\) to get the y-coordinate. If \(a > 0\), the vertex is the minimum point; if \(a < 0\), it is the maximum. This is handy for optimization problems — revenue, area, or projectile height — where the answer is the peak or bottom of the curve rather than its roots.
What are the most common mistakes when entering coefficients?
The biggest one is dropping negative signs: in \(x^2 - 3x + 2 = 0\), \(b\) is \(-3\), not 3. Others include forgetting a coefficient of 1 (in \(x^2 + 4 = 0\), \(b\) is 0, not blank), entering \(c\) with the wrong sign when it sits on the right side of the original equation, and typing \(a = 0\) for a linear equation. Always run a quick Vieta check: \(x_1 + x_2 = -\tfrac{b}{a}\) and \(x_1 x_2 = \tfrac{c}{a}\) should both hold.