Solve Using the Quadratic Formula x^2+10x=39

$x^{2} + 10 x = 39$

Step 1

Subtract $39$ from both sides of the equation.

$x^{2} + 10 x - 39 = 0$

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 $a = 1$ , $b = 10$ , and $c = - 39$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot - 39)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot - 39}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 39$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot - 39}}{2 \cdot 1}$

Multiply $- 4$ by $- 39$ .

$x = \frac{- 10 \pm \sqrt{100 + 156}}{2 \cdot 1}$

Add $100$ and $156$ .

$x = \frac{- 10 \pm \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{- 10 \pm \sqrt{16^{2}}}{2 \cdot 1}$

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

$x = \frac{- 10 \pm 16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 16}{2}$

Simplify $\frac{- 10 \pm 16}{2}$ .

$x = - 5 \pm 8$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 13$