Solve (y+10)^2-64=0

$$( y + 10 ) ^ { 2 } - 64 = 0$$

Solve for y

$$y=-2$$

$$y=-18$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+10\right)^{2}.

$$y^{2}+20y+100-64=0$$

Subtract 64 from 100 to get 36.

$$y^{2}+20y+36=0$$

To solve the equation, factor y^{2}+20y+36 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.

$$a+b=20 ab=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 20.

$$a=2 b=18$$

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

$$\left(y+2\right)\left(y+18\right)$$

To find equation solutions, solve y+2=0 and y+18=0.

$$y=-2 y=-18$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+10\right)^{2}.

$$y^{2}+20y+100-64=0$$

Subtract 64 from 100 to get 36.

$$y^{2}+20y+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+36. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\times 36=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 20.

$$a=2 b=18$$

Rewrite y^{2}+20y+36 as \left(y^{2}+2y\right)+\left(18y+36\right).

$$\left(y^{2}+2y\right)+\left(18y+36\right)$$

Factor out y in the first and 18 in the second group.

$$y\left(y+2\right)+18\left(y+2\right)$$

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

$$\left(y+2\right)\left(y+18\right)$$

To find equation solutions, solve y+2=0 and y+18=0.

$$y=-2 y=-18$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+10\right)^{2}.

$$y^{2}+20y+100-64=0$$

Subtract 64 from 100 to get 36.

$$y^{2}+20y+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-20±\sqrt{20^{2}-4\times 36}}{2}$$

Square 20.

$$y=\frac{-20±\sqrt{400-4\times 36}}{2}$$

Multiply -4 times 36.

$$y=\frac{-20±\sqrt{400-144}}{2}$$

Add 400 to -144.

$$y=\frac{-20±\sqrt{256}}{2}$$

Take the square root of 256.

$$y=\frac{-20±16}{2}$$

Now solve the equation y=\frac{-20±16}{2} when ± is plus. Add -20 to 16.

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

Divide -4 by 2.

$$y=-2$$

Now solve the equation y=\frac{-20±16}{2} when ± is minus. Subtract 16 from -20.

$$y=-\frac{36}{2}$$

Divide -36 by 2.

$$y=-18$$

The equation is now solved.

$$y=-2 y=-18$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+10\right)^{2}.

$$y^{2}+20y+100-64=0$$

Subtract 64 from 100 to get 36.

$$y^{2}+20y+36=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}+20y=-36$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+20y+10^{2}=-36+10^{2}$$

Square 10.

$$y^{2}+20y+100=-36+100$$

Add -36 to 100.

$$y^{2}+20y+100=64$$

Factor y^{2}+20y+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+10\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+10\right)^{2}}=\sqrt{64}$$

Simplify.

$$y+10=8 y+10=-8$$

Subtract 10 from both sides of the equation.

$$y=-2 y=-18$$