Solve Using the Quadratic Formula 4x^2-3x-16=-2x^2-2x-4

$4 x^{2} - 3 x - 16 = - 2 x^{2} - 2 x - 4$

Step 1

Move all terms to the left side of the equation and simplify.

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

Add $2 x^{2}$ to both sides of the equation.

$4 x^{2} - 3 x - 16 + 2 x^{2} = - 2 x - 4$

Add $2 x$ to both sides of the equation.

$4 x^{2} - 3 x - 16 + 2 x^{2} + 2 x = - 4$

Add $4$ to both sides of the equation.

$4 x^{2} - 3 x - 16 + 2 x^{2} + 2 x + 4 = 0$

Simplify $4 x^{2} - 3 x - 16 + 2 x^{2} + 2 x + 4$ .

Add $4 x^{2}$ and $2 x^{2}$ .

$6 x^{2} - 3 x - 16 + 2 x + 4 = 0$

Add $- 3 x$ and $2 x$ .

$6 x^{2} - x - 16 + 4 = 0$

Add $- 16$ and $4$ .

$6 x^{2} - x - 12 = 0$

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 $a = 6$ , $b = - 1$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (6 \cdot - 12)}}{2 \cdot 6}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 6 \cdot - 12}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 12$ .

Multiply $- 4$ by $6$ .

$x = \frac{1 \pm \sqrt{1 - 24 \cdot - 12}}{2 \cdot 6}$

Multiply $- 24$ by $- 12$ .

$x = \frac{1 \pm \sqrt{1 + 288}}{2 \cdot 6}$

Add $1$ and $288$ .

$x = \frac{1 \pm \sqrt{289}}{2 \cdot 6}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{1 \pm \sqrt{17^{2}}}{2 \cdot 6}$

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

$x = \frac{1 \pm 17}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{1 \pm 17}{12}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{3}{2}, - \frac{4}{3}$