Solve Using the Quadratic Formula x^2+11x=16+5x

$x^{2} + 11 x = 16 + 5 x$

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

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

Subtract $16$ from both sides of the equation.

$x^{2} + 11 x - 16 = 5 x$

Subtract $5 x$ from both sides of the equation.

$x^{2} + 11 x - 16 - 5 x = 0$

Subtract $5 x$ from $11 x$ .

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 6$ , and $c = - 16$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot - 16)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 16}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 16}}{2 \cdot 1}$

Multiply $- 4$ by $- 16$ .

$x = \frac{- 6 \pm \sqrt{36 + 64}}{2 \cdot 1}$

Add $36$ and $64$ .

$x = \frac{- 6 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 6 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 6 \pm 10}{2}$

Simplify $\frac{- 6 \pm 10}{2}$ .

$x = - 3 \pm 5$

The final answer is the combination of both solutions.

$x = 2, - 8$