Solve Using the Quadratic Formula 4x^2-20x+24=0

$4 x^{2} - 20 x + 24 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 4$ , $b = - 20$ , and $c = 24$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (4 \cdot 24)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 20$ to the power of $2$ .

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 4 \cdot 24}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 24$ .

Multiply $- 4$ by $4$ .

$x = \frac{20 \pm \sqrt{400 - 16 \cdot 24}}{2 \cdot 4}$

Multiply $- 16$ by $24$ .

$x = \frac{20 \pm \sqrt{400 - 384}}{2 \cdot 4}$

Subtract $384$ from $400$ .

$x = \frac{20 \pm \sqrt{16}}{2 \cdot 4}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{20 \pm \sqrt{4^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{20 \pm 4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{20 \pm 4}{8}$

Simplify $\frac{20 \pm 4}{8}$ .

$x = \frac{5 \pm 1}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 3, 2$