Solve (y+1)^2=3^2

$$( y + 1 ) ^ { 2 } = 3 ^ { 2 }$$

Solve for y

$$y=2$$

$$y=-4$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$y^{2}+2y+1=9$$

Subtract 9 from both sides.

$$y^{2}+2y+1-9=0$$

Subtract 9 from 1 to get -8.

$$y^{2}+2y-8=0$$

To solve the equation, factor y^{2}+2y-8 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.

$$a+b=2 ab=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

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

$$\left(y-2\right)\left(y+4\right)$$

To find equation solutions, solve y-2=0 and y+4=0.

$$y=2 y=-4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$y^{2}+2y+1=9$$

Subtract 9 from both sides.

$$y^{2}+2y+1-9=0$$

Subtract 9 from 1 to get -8.

$$y^{2}+2y-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite y^{2}+2y-8 as \left(y^{2}-2y\right)+\left(4y-8\right).

$$\left(y^{2}-2y\right)+\left(4y-8\right)$$

Factor out y in the first and 4 in the second group.

$$y\left(y-2\right)+4\left(y-2\right)$$

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

$$\left(y-2\right)\left(y+4\right)$$

To find equation solutions, solve y-2=0 and y+4=0.

$$y=2 y=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$y^{2}+2y+1=9$$

Subtract 9 from both sides.

$$y^{2}+2y+1-9=0$$

Subtract 9 from 1 to get -8.

$$y^{2}+2y-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-8\right)}}{2}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$y=\frac{-2±\sqrt{4+32}}{2}$$

Add 4 to 32.

$$y=\frac{-2±\sqrt{36}}{2}$$

Take the square root of 36.

$$y=\frac{-2±6}{2}$$

Now solve the equation y=\frac{-2±6}{2} when ± is plus. Add -2 to 6.

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

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-2±6}{2} when ± is minus. Subtract 6 from -2.

$$y=-\frac{8}{2}$$

Divide -8 by 2.

$$y=-4$$

The equation is now solved.

$$y=2 y=-4$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$y^{2}+2y+1=9$$

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

$$\left(y+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$y+1=3 y+1=-3$$

Subtract 1 from both sides of the equation.

$$y=2 y=-4$$