Solve Using the Quadratic Formula 4x^2+4x=19

$4 x^{2} + 4 x = 19$

Step 1

Subtract $19$ from both sides of the equation.

$4 x^{2} + 4 x - 19 = 0$

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 $a = 4$ , $b = 4$ , and $c = - 19$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (4 \cdot - 19)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 19}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 19$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 4 \pm \sqrt{16 - 16 \cdot - 19}}{2 \cdot 4}$

Multiply $- 16$ by $- 19$ .

$x = \frac{- 4 \pm \sqrt{16 + 304}}{2 \cdot 4}$

Add $16$ and $304$ .

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

Rewrite $320$ as $8^{2} \cdot 5$ .

Factor $64$ out of $320$ .

$x = \frac{- 4 \pm \sqrt{64 (5)}}{2 \cdot 4}$

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

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 5}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 8 \sqrt{5}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 4 \pm 8 \sqrt{5}}{8}$

Simplify $\frac{- 4 \pm 8 \sqrt{5}}{8}$ .

$x = \frac{- 1 \pm 2 \sqrt{5}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm 2 \sqrt{5}}{2}$

Decimal Form:

$x = 1.73606797 \dots, - 2.73606797 \dots$