Solve for x x^2+x=3

$x^{2} + x = 3$

Step 1

Subtract $3$ from both sides of the equation.

$x^{2} + x - 3 = 0$

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 $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 3)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 1 \pm \sqrt{1 + 12}}{2 \cdot 1}$

Add $1$ and $12$ .

$x = \frac{- 1 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{13}}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{13}}{2}, - \frac{1 + \sqrt{13}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{13}}{2}, - \frac{1 + \sqrt{13}}{2}$

Decimal Form:

$x = 1.30277563 \dots, - 2.30277563 \dots$