Solve left(y+2right)^2=-28(0-7)

$${ \left(y+2 \right) }^{ 2 } =-28(0-7)$$

Solve for y

$$y=12$$

$$y=-16$$

Steps Using Factoring

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

$$y^{2}+4y+4=-28\left(-7\right)$$

Multiply -28 and -7 to get 196.

$$y^{2}+4y+4=196$$

Subtract 196 from both sides.

$$y^{2}+4y+4-196=0$$

Subtract 196 from 4 to get -192.

$$y^{2}+4y-192=0$$

To solve the equation, factor y^{2}+4y-192 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.

$$a+b=4 ab=-192$$

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 -192.

$$-1,192 -2,96 -3,64 -4,48 -6,32 -8,24 -12,16$$

Calculate the sum for each pair.

$$-1+192=191 -2+96=94 -3+64=61 -4+48=44 -6+32=26 -8+24=16 -12+16=4$$

The solution is the pair that gives sum 4.

$$a=-12 b=16$$

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

$$\left(y-12\right)\left(y+16\right)$$

To find equation solutions, solve y-12=0 and y+16=0.

$$y=12 y=-16$$

Steps Using Factoring By Grouping

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

$$y^{2}+4y+4=-28\left(-7\right)$$

Multiply -28 and -7 to get 196.

$$y^{2}+4y+4=196$$

Subtract 196 from both sides.

$$y^{2}+4y+4-196=0$$

Subtract 196 from 4 to get -192.

$$y^{2}+4y-192=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-192. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-192\right)=-192$$

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 -192.

$$-1,192 -2,96 -3,64 -4,48 -6,32 -8,24 -12,16$$

Calculate the sum for each pair.

$$-1+192=191 -2+96=94 -3+64=61 -4+48=44 -6+32=26 -8+24=16 -12+16=4$$

The solution is the pair that gives sum 4.

$$a=-12 b=16$$

Rewrite y^{2}+4y-192 as \left(y^{2}-12y\right)+\left(16y-192\right).

$$\left(y^{2}-12y\right)+\left(16y-192\right)$$

Factor out y in the first and 16 in the second group.

$$y\left(y-12\right)+16\left(y-12\right)$$

Factor out common term y-12 by using distributive property.

$$\left(y-12\right)\left(y+16\right)$$

To find equation solutions, solve y-12=0 and y+16=0.

$$y=12 y=-16$$

Steps Using the Quadratic Formula

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

$$y^{2}+4y+4=-28\left(-7\right)$$

Multiply -28 and -7 to get 196.

$$y^{2}+4y+4=196$$

Subtract 196 from both sides.

$$y^{2}+4y+4-196=0$$

Subtract 196 from 4 to get -192.

$$y^{2}+4y-192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-192\right)}}{2}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-192\right)}}{2}$$

Multiply -4 times -192.

$$y=\frac{-4±\sqrt{16+768}}{2}$$

Add 16 to 768.

$$y=\frac{-4±\sqrt{784}}{2}$$

Take the square root of 784.

$$y=\frac{-4±28}{2}$$

Now solve the equation y=\frac{-4±28}{2} when ± is plus. Add -4 to 28.

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

Divide 24 by 2.

$$y=12$$

Now solve the equation y=\frac{-4±28}{2} when ± is minus. Subtract 28 from -4.

$$y=-\frac{32}{2}$$

Divide -32 by 2.

$$y=-16$$

The equation is now solved.

$$y=12 y=-16$$

Steps for Completing the Square

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

$$y^{2}+4y+4=-28\left(-7\right)$$

Multiply -28 and -7 to get 196.

$$y^{2}+4y+4=196$$

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}.

$$\left(y+2\right)^{2}=196$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{196}$$

Simplify.

$$y+2=14 y+2=-14$$

Subtract 2 from both sides of the equation.

$$y=12 y=-16$$