Solve for x square root of 5+x=x^2

$\sqrt{5} + x = x^{2}$

Step 1

Subtract $x^{2}$ from both sides of the equation.

$\sqrt{5} + x - x^{2} = 0$

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 $a = - 1$ , $b = 1$ , and $c = \sqrt{5}$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \sqrt{5}}}{- 2}$

Simplify $\frac{- 1 \pm \sqrt{1 + 4 \sqrt{5}}}{- 2}$ .

$x = \frac{1 \pm \sqrt{1 + 4 \sqrt{5}}}{2}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{1 + 4 \sqrt{5}}}{2}, \frac{1 - \sqrt{1 + 4 \sqrt{5}}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{1 + 4 \sqrt{5}}}{2}, \frac{1 - \sqrt{1 + 4 \sqrt{5}}}{2}$

Decimal Form:

$x = 2.07672698 \dots, - 1.07672698 \dots$