Solve Using the Quadratic Formula 2x^2+13x=7

$2 x^{2} + 13 x = 7$

Step 1

Subtract $7$ from both sides of the equation.

$2 x^{2} + 13 x - 7 = 0$

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 $a = 2$ , $b = 13$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 13 \pm \sqrt{13^{2} - 4 \cdot (2 \cdot - 7)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $13$ to the power of $2$ .

$x = \frac{- 13 \pm \sqrt{169 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

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

Multiply $- 4$ by $2$ .

$x = \frac{- 13 \pm \sqrt{169 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{- 13 \pm \sqrt{169 + 56}}{2 \cdot 2}$

Add $169$ and $56$ .

$x = \frac{- 13 \pm \sqrt{225}}{2 \cdot 2}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{- 13 \pm \sqrt{15^{2}}}{2 \cdot 2}$

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

$x = \frac{- 13 \pm 15}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 13 \pm 15}{4}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{1}{2}, - 7$