Solve (y+4)^2+2(y+4)y+y^2=4

$$( y + 4 ) ^ { 2 } + 2 ( y + 4 ) y + y ^ { 2 } = 4$$

Solve for y

$$y=-3$$

$$y=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.

$$y^{2}+8y+16+2\left(y+4\right)y+y^{2}=4$$

Use the distributive property to multiply 2 by y+4.

$$y^{2}+8y+16+\left(2y+8\right)y+y^{2}=4$$

Use the distributive property to multiply 2y+8 by y.

$$y^{2}+8y+16+2y^{2}+8y+y^{2}=4$$

Combine y^{2} and 2y^{2} to get 3y^{2}.

$$3y^{2}+8y+16+8y+y^{2}=4$$

Combine 8y and 8y to get 16y.

$$3y^{2}+16y+16+y^{2}=4$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$4y^{2}+16y+16=4$$

Subtract 4 from both sides.

$$4y^{2}+16y+16-4=0$$

Subtract 4 from 16 to get 12.

$$4y^{2}+16y+12=0$$

Divide both sides by 4.

$$y^{2}+4y+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite y^{2}+4y+3 as \left(y^{2}+y\right)+\left(3y+3\right).

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

Factor out y in the first and 3 in the second group.

$$y\left(y+1\right)+3\left(y+1\right)$$

Factor out common term y+1 by using distributive property.

$$\left(y+1\right)\left(y+3\right)$$

To find equation solutions, solve y+1=0 and y+3=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.

$$y^{2}+8y+16+2\left(y+4\right)y+y^{2}=4$$

Use the distributive property to multiply 2 by y+4.

$$y^{2}+8y+16+\left(2y+8\right)y+y^{2}=4$$

Use the distributive property to multiply 2y+8 by y.

$$y^{2}+8y+16+2y^{2}+8y+y^{2}=4$$

Combine y^{2} and 2y^{2} to get 3y^{2}.

$$3y^{2}+8y+16+8y+y^{2}=4$$

Combine 8y and 8y to get 16y.

$$3y^{2}+16y+16+y^{2}=4$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$4y^{2}+16y+16=4$$

Subtract 4 from both sides.

$$4y^{2}+16y+16-4=0$$

Subtract 4 from 16 to get 12.

$$4y^{2}+16y+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 16 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-16±\sqrt{16^{2}-4\times 4\times 12}}{2\times 4}$$

Square 16.

$$y=\frac{-16±\sqrt{256-4\times 4\times 12}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-16±\sqrt{256-16\times 12}}{2\times 4}$$

Multiply -16 times 12.

$$y=\frac{-16±\sqrt{256-192}}{2\times 4}$$

Add 256 to -192.

$$y=\frac{-16±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$y=\frac{-16±8}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-16±8}{8}$$

Now solve the equation y=\frac{-16±8}{8} when ± is plus. Add -16 to 8.

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

Divide -8 by 8.

$$y=-1$$

Now solve the equation y=\frac{-16±8}{8} when ± is minus. Subtract 8 from -16.

$$y=-\frac{24}{8}$$

Divide -24 by 8.

$$y=-3$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.

$$y^{2}+8y+16+2\left(y+4\right)y+y^{2}=4$$

Use the distributive property to multiply 2 by y+4.

$$y^{2}+8y+16+\left(2y+8\right)y+y^{2}=4$$

Use the distributive property to multiply 2y+8 by y.

$$y^{2}+8y+16+2y^{2}+8y+y^{2}=4$$

Combine y^{2} and 2y^{2} to get 3y^{2}.

$$3y^{2}+8y+16+8y+y^{2}=4$$

Combine 8y and 8y to get 16y.

$$3y^{2}+16y+16+y^{2}=4$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$4y^{2}+16y+16=4$$

Subtract 16 from both sides.

$$4y^{2}+16y=4-16$$

Subtract 16 from 4 to get -12.

$$4y^{2}+16y=-12$$

Divide both sides by 4.

$$\frac{4y^{2}+16y}{4}=-\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\frac{16}{4}y=-\frac{12}{4}$$

Divide 16 by 4.

$$y^{2}+4y=-\frac{12}{4}$$

Divide -12 by 4.

$$y^{2}+4y=-3$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+4y+2^{2}=-3+2^{2}$$

Square 2.

$$y^{2}+4y+4=-3+4$$

Add -3 to 4.

$$y^{2}+4y+4=1$$

Factor y^{2}+4y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y+2=1 y+2=-1$$

Subtract 2 from both sides of the equation.

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