Solve y^2+20y=4y-64

$$y ^ { 2 } + 20 y = 4 y - 64$$

Solve for y

$$y=-8$$

Steps Using Factoring

Subtract 4y from both sides.

$$y^{2}+20y-4y=-64$$

Combine 20y and -4y to get 16y.

$$y^{2}+16y=-64$$

Add 64 to both sides.

$$y^{2}+16y+64=0$$

To solve the equation, factor y^{2}+16y+64 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.

$$a+b=16 ab=64$$

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 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 16.

$$a=8 b=8$$

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

$$\left(y+8\right)\left(y+8\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve y+8=0.

$$y=-8$$

Steps Using Factoring By Grouping

Subtract 4y from both sides.

$$y^{2}+20y-4y=-64$$

Combine 20y and -4y to get 16y.

$$y^{2}+16y=-64$$

Add 64 to both sides.

$$y^{2}+16y+64=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+64. To find a and b, set up a system to be solved.

$$a+b=16 ab=1\times 64=64$$

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 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 16.

$$a=8 b=8$$

Rewrite y^{2}+16y+64 as \left(y^{2}+8y\right)+\left(8y+64\right).

$$\left(y^{2}+8y\right)+\left(8y+64\right)$$

Factor out y in the first and 8 in the second group.

$$y\left(y+8\right)+8\left(y+8\right)$$

Factor out common term y+8 by using distributive property.

$$\left(y+8\right)\left(y+8\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve y+8=0.

$$y=-8$$

Steps Using the Quadratic Formula

Subtract 4y from both sides.

$$y^{2}+20y-4y=-64$$

Combine 20y and -4y to get 16y.

$$y^{2}+16y=-64$$

Add 64 to both sides.

$$y^{2}+16y+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 16 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-16±\sqrt{16^{2}-4\times 64}}{2}$$

Square 16.

$$y=\frac{-16±\sqrt{256-4\times 64}}{2}$$

Multiply -4 times 64.

$$y=\frac{-16±\sqrt{256-256}}{2}$$

Add 256 to -256.

$$y=\frac{-16±\sqrt{0}}{2}$$

Take the square root of 0.

$$y=-\frac{16}{2}$$

Divide -16 by 2.

$$y=-8$$

Steps for Completing the Square

Subtract 4y from both sides.

$$y^{2}+20y-4y=-64$$

Combine 20y and -4y to get 16y.

$$y^{2}+16y=-64$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+16y+8^{2}=-64+8^{2}$$

Square 8.

$$y^{2}+16y+64=-64+64$$

Add -64 to 64.

$$y^{2}+16y+64=0$$

Factor y^{2}+16y+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+8\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+8\right)^{2}}=\sqrt{0}$$

Simplify.

$$y+8=0 y+8=0$$

Subtract 8 from both sides of the equation.

$$y=-8 y=-8$$

The equation is now solved. Solutions are the same.

$$y=-8$$