Solve Using the Quadratic Formula x^2+6x=x+50

$x^{2} + 6 x = x + 50$

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

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

Subtract $x$ from both sides of the equation.

$x^{2} + 6 x - x = 50$

Subtract $50$ from both sides of the equation.

$x^{2} + 6 x - x - 50 = 0$

Subtract $x$ from $6 x$ .

$x^{2} + 5 x - 50 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 5$ , and $c = - 50$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 50)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 50}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 50}}{2 \cdot 1}$

Multiply $- 4$ by $- 50$ .

$x = \frac{- 5 \pm \sqrt{25 + 200}}{2 \cdot 1}$

Add $25$ and $200$ .

$x = \frac{- 5 \pm \sqrt{225}}{2 \cdot 1}$

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

$x = \frac{- 5 \pm \sqrt{15^{2}}}{2 \cdot 1}$

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

$x = \frac{- 5 \pm 15}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm 15}{2}$

The final answer is the combination of both solutions.

$x = 5, - 10$