Solve Using the Quadratic Formula 2x^2+4x=7

$2 x^{2} + 4 x = 7$

Step 1

Subtract $7$ from both sides of the equation.

$2 x^{2} + 4 x - 7 = 0$

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 $a = 2$ , $b = 4$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (2 \cdot - 7)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{- 4 \pm \sqrt{16 + 56}}{2 \cdot 2}$

Add $16$ and $56$ .

$x = \frac{- 4 \pm \sqrt{72}}{2 \cdot 2}$

Rewrite $72$ as $6^{2} \cdot 2$ .

Factor $36$ out of $72$ .

$x = \frac{- 4 \pm \sqrt{36 (2)}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $2$ .

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

Simplify $\frac{- 4 \pm 6 \sqrt{2}}{4}$ .

$x = \frac{- 2 \pm 3 \sqrt{2}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 2 \pm 3 \sqrt{2}}{2}$

Decimal Form:

$x = 1.12132034 \dots, - 3.12132034 \dots$