Solve Using the Quadratic Formula 4x^2-6x-18=0

$4 x^{2} - 6 x - 18 = 0$

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 $a = 4$ , $b = - 6$ , and $c = - 18$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (4 \cdot - 18)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 4 \cdot - 18}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 18$ .

Multiply $- 4$ by $4$ .

$x = \frac{6 \pm \sqrt{36 - 16 \cdot - 18}}{2 \cdot 4}$

Multiply $- 16$ by $- 18$ .

$x = \frac{6 \pm \sqrt{36 + 288}}{2 \cdot 4}$

Add $36$ and $288$ .

$x = \frac{6 \pm \sqrt{324}}{2 \cdot 4}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{6 \pm \sqrt{18^{2}}}{2 \cdot 4}$

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

$x = \frac{6 \pm 18}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{6 \pm 18}{8}$

Simplify $\frac{6 \pm 18}{8}$ .

$x = \frac{3 \pm 9}{4}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - \frac{3}{2}$