Solve Using the Quadratic Formula 4x^2-32x+60=0

$4 x^{2} - 32 x + 60 = 0$

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 $a = 4$ , $b = - 32$ , and $c = 60$ into the quadratic formula and solve for $x$ .

$\frac{32 \pm \sqrt{{(- 32)}^{2} - 4 \cdot (4 \cdot 60)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 32$ to the power of $2$ .

$x = \frac{32 \pm \sqrt{1024 - 4 \cdot 4 \cdot 60}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 60$ .

Multiply $- 4$ by $4$ .

$x = \frac{32 \pm \sqrt{1024 - 16 \cdot 60}}{2 \cdot 4}$

Multiply $- 16$ by $60$ .

$x = \frac{32 \pm \sqrt{1024 - 960}}{2 \cdot 4}$

Subtract $960$ from $1024$ .

$x = \frac{32 \pm \sqrt{64}}{2 \cdot 4}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{32 \pm \sqrt{8^{2}}}{2 \cdot 4}$

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

$x = \frac{32 \pm 8}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{32 \pm 8}{8}$

Simplify $\frac{32 \pm 8}{8}$ .

$x = 4 \pm 1$

Step 4

The final answer is the combination of both solutions.

$x = 5, 3$