Solve (y+4)^2-4=-3(y+4)

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

Solve for y

$$y=-8$$

$$y=-3$$

Steps Using Factoring

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

$$y^{2}+8y+16-4=-3\left(y+4\right)$$

Subtract 4 from 16 to get 12.

$$y^{2}+8y+12=-3\left(y+4\right)$$

Use the distributive property to multiply -3 by y+4.

$$y^{2}+8y+12=-3y-12$$

Add 3y to both sides.

$$y^{2}+8y+12+3y=-12$$

Combine 8y and 3y to get 11y.

$$y^{2}+11y+12=-12$$

Add 12 to both sides.

$$y^{2}+11y+12+12=0$$

Add 12 and 12 to get 24.

$$y^{2}+11y+24=0$$

To solve the equation, factor y^{2}+11y+24 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.

$$a+b=11 ab=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

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

$$\left(y+3\right)\left(y+8\right)$$

To find equation solutions, solve y+3=0 and y+8=0.

$$y=-3 y=-8$$

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}.

$$y^{2}+8y+16-4=-3\left(y+4\right)$$

Subtract 4 from 16 to get 12.

$$y^{2}+8y+12=-3\left(y+4\right)$$

Use the distributive property to multiply -3 by y+4.

$$y^{2}+8y+12=-3y-12$$

Add 3y to both sides.

$$y^{2}+8y+12+3y=-12$$

Combine 8y and 3y to get 11y.

$$y^{2}+11y+12=-12$$

Add 12 to both sides.

$$y^{2}+11y+12+12=0$$

Add 12 and 12 to get 24.

$$y^{2}+11y+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+24. To find a and b, set up a system to be solved.

$$a+b=11 ab=1\times 24=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite y^{2}+11y+24 as \left(y^{2}+3y\right)+\left(8y+24\right).

$$\left(y^{2}+3y\right)+\left(8y+24\right)$$

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

$$y\left(y+3\right)+8\left(y+3\right)$$

Factor out common term y+3 by using distributive property.

$$\left(y+3\right)\left(y+8\right)$$

To find equation solutions, solve y+3=0 and y+8=0.

$$y=-3 y=-8$$

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}.

$$y^{2}+8y+16-4=-3\left(y+4\right)$$

Subtract 4 from 16 to get 12.

$$y^{2}+8y+12=-3\left(y+4\right)$$

Use the distributive property to multiply -3 by y+4.

$$y^{2}+8y+12=-3y-12$$

Add 3y to both sides.

$$y^{2}+8y+12+3y=-12$$

Combine 8y and 3y to get 11y.

$$y^{2}+11y+12=-12$$

Add 12 to both sides.

$$y^{2}+11y+12+12=0$$

Add 12 and 12 to get 24.

$$y^{2}+11y+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 11 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-11±\sqrt{11^{2}-4\times 24}}{2}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\times 24}}{2}$$

Multiply -4 times 24.

$$y=\frac{-11±\sqrt{121-96}}{2}$$

Add 121 to -96.

$$y=\frac{-11±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-11±5}{2}$$

Now solve the equation y=\frac{-11±5}{2} when ± is plus. Add -11 to 5.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

Now solve the equation y=\frac{-11±5}{2} when ± is minus. Subtract 5 from -11.

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

Divide -16 by 2.

$$y=-8$$

The equation is now solved.

$$y=-3 y=-8$$

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}.

$$y^{2}+8y+16-4=-3\left(y+4\right)$$

Subtract 4 from 16 to get 12.

$$y^{2}+8y+12=-3\left(y+4\right)$$

Use the distributive property to multiply -3 by y+4.

$$y^{2}+8y+12=-3y-12$$

Add 3y to both sides.

$$y^{2}+8y+12+3y=-12$$

Combine 8y and 3y to get 11y.

$$y^{2}+11y+12=-12$$

Subtract 12 from both sides.

$$y^{2}+11y=-12-12$$

Subtract 12 from -12 to get -24.

$$y^{2}+11y=-24$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+11y+\left(\frac{11}{2}\right)^{2}=-24+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+11y+\frac{121}{4}=-24+\frac{121}{4}$$

Add -24 to \frac{121}{4}.

$$y^{2}+11y+\frac{121}{4}=\frac{25}{4}$$

Factor y^{2}+11y+\frac{121}{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}.

$$\left(y+\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$y+\frac{11}{2}=\frac{5}{2} y+\frac{11}{2}=-\frac{5}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$y=-3 y=-8$$