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\left(x+4\right)\left(x+4\right)=xx
Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x,x+4.
\left(x+4\right)^{2}=xx
Multiply x+4 and x+4 to get \left(x+4\right)^{2}.
\left(x+4\right)^{2}=x^{2}
Multiply x and x to get x^{2}.
x^{2}+8x+16=x^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.
x^{2}+8x+16-x^{2}=0
Subtract x^{2} from both sides.
8x+16=0
Combine x^{2} and -x^{2} to get 0.
8x=-16
Subtract 16 from both sides. Anything subtracted from zero gives its negation.
x=\frac{-16}{8}
Divide both sides by 8.
x=-2
Divide -16 by 8 to get -2.