Quadratic Formula Calculator

Using a Quadratic Formula Calculator

Our quadratic formula calculator is a highly efficient, easy-to-use tool designed to instantly solve quadratic equations. In algebra, a quadratic equation is any second-degree polynomial equation that can be written in the standard form:

ax²+bx+c=0

where

a≠0

To use this step-by-step quadratic equation solver, simply input the coefficients A, B, and C into their respective fields and click "Calculate." Please note that A cannot equal zero, whereas zero is an entirely acceptable input for B and C. Whether your equation has real or complex roots, the calculator applies the quadratic formula to find all possible solutions. Furthermore, it automatically simplifies the resulting radicals, delivering the final answers in their most reduced and accurate form.

Solving Quadratic Equations Using the Quadratic Formula

The quadratic formula is a universal method that allows you to solve any quadratic equation. To use this method, you must first arrange your given equation into the standard form: ax²+bx+c=0. From there, the exact solutions can be calculated using the following equation:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}$$

The expression beneath the square root, b²-4ac, is known as the discriminant. It is a critical value that determines the nature of the roots:

• If the discriminant is positive, b²-4ac>0, the equation will yield two distinct real roots.
• If the discriminant is negative, b²-4ac<0, the equation will produce two complex roots, since taking the square root of a negative number results in an imaginary number.
• If the discriminant equals zero, b²-4ac=0, the equation will have exactly one real root (a repeated root).

Our quadratic calculator doesn't just display the final answers; it provides the complete, step-by-step workflow for finding these solutions. It also computes the discriminant to clearly demonstrate whether it is positive, negative, or equal to zero.

Practical Examples

Example 1 (with real roots)

Let's solve the following quadratic equation:

2x²+3x-2=0

In this example

a=2,b=3,c=-2.

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-3±\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3±\sqrt{9+16}}{4}=\frac{-3±\sqrt{25}}{4}$$

The discriminant of this equation is positive,

b²-4ac=25>0

Therefore, the equation will have two real roots.

Now let's simplify the resulting radical:

$$x=\frac{-3±\sqrt{25}}{4}=\frac{-3±5}{4}$$

$$x=\frac{-3+5}{4}\ \ \ and\ \ \ x= \frac{-3-5}{4}$$

$$x=\frac{2}{4}\ \ \ and\ \ \ x=-\frac{8}{4}$$

$$x=\frac{1}{2}\ \ \ and\ \ \ x=-2$$

Finally

x=0.5

x=-2

Example 2 (with complex roots)

Let's solve the following quadratic equation:

x²+2x+5=0

In this example

a=1,b=2,c=5

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-2±\sqrt{2^2-4(1)(5)}}{2(1)}=\frac{-2±\sqrt{4-20}}{2}=\frac{-2±\sqrt{-16}}{2}$$

The discriminant of this equation is negative,

b²-4ac=-16<0

Therefore, the equation will have two complex roots.

Now let's simplify the resulting radical:

$$x=\frac{-2±\sqrt{-16}}{2}=\frac{-2±4i}{2}=\frac{-2}{2}±\frac{4i}{2}=-1±2i$$

Finally,

x=-1+2i

x=-1-2i

Example 3 (with one root)

Let's solve the following quadratic equation:

3x²+6x+3=0

In this example

a=3,b=6,c=3

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-6±\sqrt{6^2-4(3)(3)}}{2(3)}=\frac{-6±\sqrt{36-36}}{6}=\frac{-6±\sqrt0}{6}$$

The discriminant of this equation equals zero, b²-4ac=0. Therefore, the equation will have exactly one root.

$$x=\frac{-6}{6}$$

Finally,

x=-1

Derivation of the Quadratic Formula

As demonstrated in the examples above, you can confidently use the quadratic formula to solve absolutely any quadratic equation, regardless of whether the discriminant is positive, negative, or zero. But where does this formula come from? Understanding the fundamental principles of its derivation is incredibly helpful, especially if you ever forget the formula itself.

The derivation process is relatively straightforward and relies on a classic algebraic technique known as "completing the square." To derive the roots of the standard quadratic equation ax²+bx+c=0, follow these systematic steps:

1. We begin with the standard equation:

ax²+bx+c=0

Move the constant C to the right side of the equation:

ax²+bx=-c

2. Eliminate the coefficient A next to the squared term x². To do this, divide the entire equation by A:

$$x²+\frac{b}{a}x=-\frac{c}{a}$$

3. Add

$$(\frac{b}{2a})^2$$

to both sides of the equation:

$$x²+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$$

4. The left-hand side now forms a perfect square trinomial in the form

x²+2dx+d²

This expression can be conveniently rewritten as

(x+d)²

In our equation, d is expressed as

$$\frac{b}{2a}$$

So:

$$x²+\frac{b}{a}x+(\frac{b}{2a})^2 = \left(x+\frac{b}{2a}\right)^2$$

Substitute this back into the left-hand side of our formula, leaving the right-hand side untouched for now:

$$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+(\frac{b}{2a})^2$$

Now, the variable x appears only once in the entire equation.

5. Extract the square root from both sides of the equation:

$$x+\frac{b}{2a}=± \sqrt{-\frac{c}{a}+(\frac{b}{2a})^2}$$

6. Isolate x by moving $\frac{b}{2a}$ to the right side of the equation:

$$x=-\frac{b}{2a}± \sqrt{-\frac{c}{a}+(\frac{b}{2a})^2}$$

7. Multiply the right side of the equation by

$$\frac{2a}{2a}$$

$$x=\frac{-b ± \sqrt{-\frac{c}{a} × (2a)^2 + (\frac{b}{2a})^2 × (2a)^2}}{2a}$$

8. Simplify the resulting expression:

$$x=\frac{-b±\sqrt{-4ac+b²}}{2a}$$

9. As a result, we arrive at the standard quadratic formula:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}$$

Interesting Facts About Quadratic Equations

• The sum of the two roots of a quadratic equation is always equal to

$$\frac{-b}{a}$$

Consequently, if the discriminant b²-4ac equals zero, you can quickly find the single repeated root of the equation using

$$\frac{-b}{2a}$$

• The product of the two roots of a quadratic equation is exactly equal to

$$\frac{c}{a}$$

• The term "quadratic" originates from the Latin word quadratus, which translates to "square." The equation earned this name because the highest power of the variable is 2, meaning the leading variable is "squared."

• The quadratic formula in its present shape was documented as early as 628 AD by the brilliant Indian mathematician Brahmagupta. Interestingly, he didn't use modern symbols; instead, he explained the mathematical solution entirely in words. Brahmagupta also detailed only one of the two possible solutions, omitting the crucial ± sign before the square root.

• The graphical representation of a quadratic function y=ax²+bx+c forms a curved shape known as a parabola. The solutions, or roots, of the quadratic equation represent the exact coordinates where the parabola intersects the x-axis (x-intercepts). If the equation has two real roots, the graph crosses the x-axis twice. If there is only one real root, the vertex of the parabola simply touches the x-axis at its maximum or minimum point. If the equation has complex roots, the parabola never intersects the x-axis at all.

• As the value of the leading coefficient, A, approaches zero, the graph of the corresponding parabola becomes progressively flatter, eventually tending toward a straight line. Naturally, when a=0, the equation simply drops to a linear equation, and its graph becomes a perfectly straight line!

• The coefficient A also dictates the parabola's overall direction. When a>0, the parabola opens upwards into a "U" shape. Conversely, if a<0, the parabola opens downwards. And as mentioned, if a=0, the "parabola" flattens completely into a linear straight line.

Quadratic equations are widely utilized across all scientific disciplines. In physics, for instance, they are essential mathematical tools used to calculate trajectories, model kinematics, and accurately describe projectile motion.