Solve for x x^2+x=1

$x^{2} + x = 1$

Subtract $1$ from both sides of the equation.

$x^{2} + x - 1 = 0$

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

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Simplify the numerator.

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One to any power is one.

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $1$ and $4$ .

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

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.61803398 \dots, - 1.61803398 \dots$