Solve for x y=x^2+x

$y = x^{2} + x$

Step 1

Rewrite the equation as $x^{2} + x = y$ .

$x^{2} + x = y$

Step 2

Subtract $y$ from both sides of the equation.

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

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

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

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

Step 5

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Multiply $2$ by $1$ .

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

Step 6

The final answer is the combination of both solutions.

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

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