The Quadratic Formula is way of finding the solutions/roots of a quadratic function where $$a$$ ≠ 0. This works for any quadratic function set in Standard Form ($$ax^2 + bx + c = 0$$).

This formula can be expressed as:

$$x = \cfrac{-\textcolor{green}b \pm \sqrt{\textcolor{green}b^2 - 4\textcolor{red}a\textcolor{blue}c}}{2\textcolor{red}a}$$

• $$x$$ represents the roots
• $$\textcolor{red}a$$ represents the coefficient of $$x^2$$
• $$\textcolor{green}b$$ represents the coefficient of $$x$$
• $$\textcolor{blue}c$$ represents the constant

Example

Determine the roots for the quadratic function $$2x^2 + 7x - 4 = 0$$.

Since we can identify $$\textcolor{red}{a = 2}$$, $$\textcolor{green}{b = 7}$$, and $$\textcolor{blue}{c= -4}$$, we can plug these values into the quadratic formula to solve for the roots:

$$x = \cfrac{-\textcolor{green}b \pm \sqrt{\textcolor{green}b^2 - 4\textcolor{red}a\textcolor{blue}c}}{2\textcolor{red}a}$$

$$x = \cfrac{-\textcolor{green}{7} \pm \sqrt{(\textcolor{green}{7})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{-4})}}{2(\textcolor{red}{2})}$$

$$x = \cfrac{-7 \pm \sqrt{49 + 32}}{4}$$

$$x = \cfrac{-7 \pm \sqrt{81}}{4}$$

$$x = \cfrac{-7 \pm 9}{4}$$

Using ±, we can determine the 2 roots either by adding the 2 values of subtracting the 2 values:

$$x₁ = \cfrac{-7 - 9}{4}$$
$$x₁ = \cfrac{-16}{4}$$
$$x₁ = -4$$

$$x₂ = \cfrac{-7 + 9}{4}$$
$$x₂ = \cfrac{2}{4}$$
$$x₂ = \cfrac{1}{2}$$

Therefore, we can determine that the roots are $$-4$$ and $$\cfrac{1}{2}$$.

Enter in coefficients for the trinomial or click on the button to generate a new question. Then, use the quadratic equation to solve for the roots.

Determine the roots for the quadratic formula $$x(2x - 3) = 7$$.

## Number of Solutions (Discriminant)

The discriminant is the part of the quadratic formula represented as $$\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}$$. The value of the discriminant is used to determine how many roots/solutions the quadratic function has:

• When $$\textcolor{green}{b}^2 - 4\textcolor{red}a\textcolor{blue}{c} = 0$$, there is 1 real solution
• When $$\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c} > 0$$, there are 2 real solutions
• When $$\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c} < 0$$, there are 0 real solutions or 2 imaginary solutions

If there is only 1 real solution, this can be referred to as a double root. This is where $$r$$ and $$s$$ in $$a(x - r)(x - s)$$ are the same value.
If there are no real solutions, this results in 2 imaginary solutions that can only be identified by using the quadratic formula instead of factoring.

Determine how many roots the following quadratic functions have:

$$x^2 + 5x + 4$$

$$3x^2 + -7x + 5$$

$$-x^2 + 6x - 9$$

The hypoteneuse of a triangle is $$20\;[cm]$$. The sum of the lengths of the legs is $$28\;[cm]$$. Find the length of each leg.