Solve -3(y+4)^2=-192

$$- 3 ( y + 4 ) ^ { 2 } = - 192$$

Solve for y

$$y=4$$

$$y=-12$$

Steps Using Factoring

Divide both sides by -3.

$$\left(y+4\right)^{2}=\frac{-192}{-3}$$

Divide -192 by -3 to get 64.

$$\left(y+4\right)^{2}=64$$

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

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

Subtract 64 from both sides.

$$y^{2}+8y+16-64=0$$

Subtract 64 from 16 to get -48.

$$y^{2}+8y-48=0$$

To solve the equation, factor y^{2}+8y-48 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.

$$a+b=8 ab=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

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

$$\left(y-4\right)\left(y+12\right)$$

To find equation solutions, solve y-4=0 and y+12=0.

$$y=4 y=-12$$

Steps Using Factoring By Grouping

Divide both sides by -3.

$$\left(y+4\right)^{2}=\frac{-192}{-3}$$

Divide -192 by -3 to get 64.

$$\left(y+4\right)^{2}=64$$

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

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

Subtract 64 from both sides.

$$y^{2}+8y+16-64=0$$

Subtract 64 from 16 to get -48.

$$y^{2}+8y-48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-48. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-48\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite y^{2}+8y-48 as \left(y^{2}-4y\right)+\left(12y-48\right).

$$\left(y^{2}-4y\right)+\left(12y-48\right)$$

Factor out y in the first and 12 in the second group.

$$y\left(y-4\right)+12\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y+12\right)$$

To find equation solutions, solve y-4=0 and y+12=0.

$$y=4 y=-12$$

Steps Using the Quadratic Formula

Divide both sides by -3.

$$\left(y+4\right)^{2}=\frac{-192}{-3}$$

Divide -192 by -3 to get 64.

$$\left(y+4\right)^{2}=64$$

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

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

Subtract 64 from both sides.

$$y^{2}+8y+16-64=0$$

Subtract 64 from 16 to get -48.

$$y^{2}+8y-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\left(-48\right)}}{2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\left(-48\right)}}{2}$$

Multiply -4 times -48.

$$y=\frac{-8±\sqrt{64+192}}{2}$$

Add 64 to 192.

$$y=\frac{-8±\sqrt{256}}{2}$$

Take the square root of 256.

$$y=\frac{-8±16}{2}$$

Now solve the equation y=\frac{-8±16}{2} when ± is plus. Add -8 to 16.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{-8±16}{2} when ± is minus. Subtract 16 from -8.

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

Divide -24 by 2.

$$y=-12$$

The equation is now solved.

$$y=4 y=-12$$

Steps for Completing the Square

Divide both sides by -3.

$$\left(y+4\right)^{2}=\frac{-192}{-3}$$

Divide -192 by -3 to get 64.

$$\left(y+4\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+4\right)^{2}}=\sqrt{64}$$

Simplify.

$$y+4=8 y+4=-8$$

Subtract 4 from both sides of the equation.

$$y=4 y=-12$$