Solve Using the Quadratic Formula 4x^2-18x+20=0

$4 x^{2} - 18 x + 20 = 0$

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 $a = 4$ , $b = - 18$ , and $c = 20$ into the quadratic formula and solve for $x$ .

$\frac{18 \pm \sqrt{{(- 18)}^{2} - 4 \cdot (4 \cdot 20)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 18$ to the power of $2$ .

$x = \frac{18 \pm \sqrt{324 - 4 \cdot 4 \cdot 20}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 20$ .

Multiply $- 4$ by $4$ .

$x = \frac{18 \pm \sqrt{324 - 16 \cdot 20}}{2 \cdot 4}$

Multiply $- 16$ by $20$ .

$x = \frac{18 \pm \sqrt{324 - 320}}{2 \cdot 4}$

Subtract $320$ from $324$ .

$x = \frac{18 \pm \sqrt{4}}{2 \cdot 4}$

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

$x = \frac{18 \pm \sqrt{2^{2}}}{2 \cdot 4}$

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

$x = \frac{18 \pm 2}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{18 \pm 2}{8}$

Simplify $\frac{18 \pm 2}{8}$ .

$x = \frac{9 \pm 1}{4}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{2}, 2$