\(5x+x^2-15=3x\\ \Leftrightarrow x^2+2x-15=0\\ \Leftrightarrow\left(x^2+2x+1\right)-16=0\\ \Leftrightarrow\left(x+1\right)^2-4^2=0\\ \Leftrightarrow\left(x+1-4\right)\left(x+1+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
5x+x2-15=3x
5x+x2-15-3x=0
2x+x2-15=0
x2+5x-3x-15=0
x(x+5)-3(x+5)=0
(x-3)(x+5)=0
=> x ∈ {3; -5 }