Solve Using the Quadratic Formula 4x^2=14x+8

$4 x^{2} = 14 x + 8$

Step 1

Move all the expressions to the left side of the equation.

Subtract $14 x$ from both sides of the equation.

$4 x^{2} - 14 x = 8$

Subtract $8$ from both sides of the equation.

$4 x^{2} - 14 x - 8 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = - 14$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (4 \cdot - 8)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$x = \frac{14 \pm \sqrt{196 - 4 \cdot 4 \cdot - 8}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 8$ .

Multiply $- 4$ by $4$ .

$x = \frac{14 \pm \sqrt{196 - 16 \cdot - 8}}{2 \cdot 4}$

Multiply $- 16$ by $- 8$ .

$x = \frac{14 \pm \sqrt{196 + 128}}{2 \cdot 4}$

Add $196$ and $128$ .

$x = \frac{14 \pm \sqrt{324}}{2 \cdot 4}$

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

$x = \frac{14 \pm \sqrt{18^{2}}}{2 \cdot 4}$

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

$x = \frac{14 \pm 18}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{14 \pm 18}{8}$

Simplify $\frac{14 \pm 18}{8}$ .

$x = \frac{7 \pm 9}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - \frac{1}{2}$