Solve left(y+3right)^2=81

$${ \left(y+3 \right) }^{ 2 } =81$$

Solve for y

$$y=6$$

$$y=-12$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=81$$

Subtract 81 from both sides.

$$y^{2}+6y+9-81=0$$

Subtract 81 from 9 to get -72.

$$y^{2}+6y-72=0$$

To solve the equation, factor y^{2}+6y-72 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.

$$a+b=6 ab=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

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

$$\left(y-6\right)\left(y+12\right)$$

To find equation solutions, solve y-6=0 and y+12=0.

$$y=6 y=-12$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=81$$

Subtract 81 from both sides.

$$y^{2}+6y+9-81=0$$

Subtract 81 from 9 to get -72.

$$y^{2}+6y-72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-72. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-72\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite y^{2}+6y-72 as \left(y^{2}-6y\right)+\left(12y-72\right).

$$\left(y^{2}-6y\right)+\left(12y-72\right)$$

Factor out y in the first and 12 in the second group.

$$y\left(y-6\right)+12\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(y+12\right)$$

To find equation solutions, solve y-6=0 and y+12=0.

$$y=6 y=-12$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=81$$

Subtract 81 from both sides.

$$y^{2}+6y+9-81=0$$

Subtract 81 from 9 to get -72.

$$y^{2}+6y-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\left(-72\right)}}{2}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\left(-72\right)}}{2}$$

Multiply -4 times -72.

$$y=\frac{-6±\sqrt{36+288}}{2}$$

Add 36 to 288.

$$y=\frac{-6±\sqrt{324}}{2}$$

Take the square root of 324.

$$y=\frac{-6±18}{2}$$

Now solve the equation y=\frac{-6±18}{2} when ± is plus. Add -6 to 18.

$$y=\frac{12}{2}$$

Divide 12 by 2.

$$y=6$$

Now solve the equation y=\frac{-6±18}{2} when ± is minus. Subtract 18 from -6.

$$y=-\frac{24}{2}$$

Divide -24 by 2.

$$y=-12$$

The equation is now solved.

$$y=6 y=-12$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(y+3\right)^{2}}=\sqrt{81}$$

Simplify.

$$y+3=9 y+3=-9$$

Subtract 3 from both sides of the equation.

$$y=6 y=-12$$