Solve y^2=2+y

$$y ^ { 2 } = 2 + y$$

Solve for y

$$y=-1$$

$$y=2$$

Steps Using Factoring

Subtract 2 from both sides.

$$y^{2}-2=y$$

Subtract y from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor y^{2}-y-2 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-2\right)\left(y+1\right)$$

To find equation solutions, solve y-2=0 and y+1=0.

$$y=2 y=-1$$

Steps Using Factoring By Grouping

Subtract 2 from both sides.

$$y^{2}-2=y$$

Subtract y from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite y^{2}-y-2 as \left(y^{2}-2y\right)+\left(y-2\right).

$$\left(y^{2}-2y\right)+\left(y-2\right)$$

Factor out y in y^{2}-2y.

$$y\left(y-2\right)+y-2$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y+1\right)$$

To find equation solutions, solve y-2=0 and y+1=0.

$$y=2 y=-1$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

$$y^{2}-2=y$$

Subtract y from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$y=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$y=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$y=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$y=\frac{1±3}{2}$$

Now solve the equation y=\frac{1±3}{2} when ± is plus. Add 1 to 3.

$$y=\frac{4}{2}$$

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

$$y=-\frac{2}{2}$$

Divide -2 by 2.

$$y=-1$$

The equation is now solved.

$$y=2 y=-1$$

Steps for Completing the Square

Subtract y from both sides.

$$y^{2}-y=2$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=2+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$y^{2}-y+\frac{1}{4}=\frac{9}{4}$$

Factor y^{2}-y+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{3}{2} y-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$y=2 y=-1$$