Solve Using the Quadratic Formula x^2-6x-7=2x

$x^{2} - 6 x - 7 = 2 x$

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

Subtract $2 x$ from both sides of the equation.

$x^{2} - 6 x - 7 - 2 x = 0$

Subtract $2 x$ from $- 6 x$ .

$x^{2} - 8 x - 7 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 8$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot - 7)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$x = \frac{8 \pm \sqrt{64 + 28}}{2 \cdot 1}$

Add $64$ and $28$ .

$x = \frac{8 \pm \sqrt{92}}{2 \cdot 1}$

Rewrite $92$ as $2^{2} \cdot 23$ .

Factor $4$ out of $92$ .

$x = \frac{8 \pm \sqrt{4 (23)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{8 \pm \sqrt{2^{2} \cdot 23}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{23}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{23}}{2}$

Simplify $\frac{8 \pm 2 \sqrt{23}}{2}$ .

$x = 4 \pm \sqrt{23}$

The result can be shown in multiple forms.

Exact Form:

$x = 4 \pm \sqrt{23}$

Decimal Form:

$x = 8.79583152 \dots, - 0.79583152 \dots$