Solve for x x^2=23+2x

$x^{2} = 23 + 2 x$

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

$x^{2} - 2 x = 23$

Subtract $23$ from both sides of the equation.

$x^{2} - 2 x - 23 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 23$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 23)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 23}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 23}}{2 \cdot 1}$

Multiply $- 4$ by $- 23$ .

$x = \frac{2 \pm \sqrt{4 + 92}}{2 \cdot 1}$

Add $4$ and $92$ .

$x = \frac{2 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{2 \pm \sqrt{16 (6)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{2 \pm \sqrt{4^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

$x = \frac{2 \pm 4 \sqrt{6}}{2}$

Simplify $\frac{2 \pm 4 \sqrt{6}}{2}$ .

$x = 1 \pm 2 \sqrt{6}$

The final answer is the combination of both solutions.

$x = 1 + 2 \sqrt{6}, 1 - 2 \sqrt{6}$

The result can be shown in multiple forms.

Exact Form:

$x = 1 + 2 \sqrt{6}, 1 - 2 \sqrt{6}$

Decimal Form:

$x = 5.89897948 \dots, - 3.89897948 \dots$