Question

(x+1)² +(5 -x)(x+5) = 0

Answer 1

\(\left(x+1\right)^2+\left(5-x\right)\left(x+5\right)=0\\ x^2+2x+1+25-x^2=0\\ x^2-x^2+2x+1+25=0\\ 2x+26=0\\ 2x=0-26\\ 2x=-26\\ x=-26:2\\ x=-13\)

Answer 2

\(\left(x+1\right)^2+\left(5-x\right)\left(x+5\right)=0\)

\(x^2+2x+1+\left(5x+25-x^2-5x\right)=0\)

\(x^2+2x+1+\left[\left(5x-5x\right)+25-x^2\right]=0\)

\(x^2+2x+1+25-x^2=0\)

\(=\left(x^2-x^2\right)+2x+1+25=0\)

\(2x+26=0\)

\(2x\)         \(=0-26=-26\)

  \(x\)         \(=\left(-26\right):2=-13\)