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\sqrt{y+4}=y+4
Subtract -4 from both sides of the equation.
\left(\sqrt{y+4}\right)^{2}=\left(y+4\right)^{2}
Square both sides of the equation.
y+4=\left(y+4\right)^{2}
Calculate \sqrt{y+4} to the power of 2 and get y+4.
y+4=y^{2}+8y+16
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+4\right)^{2}.
y+4-y^{2}=8y+16
Subtract y^{2} from both sides.
y+4-y^{2}-8y=16
Subtract 8y from both sides.
-7y+4-y^{2}=16
Combine y and -8y to get -7y.
-7y+4-y^{2}-16=0
Subtract 16 from both sides.
-7y-12-y^{2}=0
Subtract 16 from 4 to get -12.
-y^{2}-7y-12=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-7 ab=-\left(-12\right)=12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-12. To find a and b, set up a system to be solved.
-1,-12 -2,-6 -3,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.
-1-12=-13 -2-6=-8 -3-4=-7
Calculate the sum for each pair.
a=-3 b=-4
The solution is the pair that gives sum -7.
\left(-y^{2}-3y\right)+\left(-4y-12\right)
Rewrite -y^{2}-7y-12 as \left(-y^{2}-3y\right)+\left(-4y-12\right).
y\left(-y-3\right)+4\left(-y-3\right)
Factor out y in the first and 4 in the second group.
\left(-y-3\right)\left(y+4\right)
Factor out common term -y-3 by using distributive property.
y=-3 y=-4
To find equation solutions, solve -y-3=0 and y+4=0.
\sqrt{-3+4}-4=-3
Substitute -3 for y in the equation \sqrt{y+4}-4=y.
-3=-3
Simplify. The value y=-3 satisfies the equation.
\sqrt{-4+4}-4=-4
Substitute -4 for y in the equation \sqrt{y+4}-4=y.
-4=-4
Simplify. The value y=-4 satisfies the equation.
y=-3 y=-4
List all solutions of \sqrt{y+4}=y+4.