Question

tim x . 
A.     2(x +5)-x^2-5x =0

Answer 1

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-\left(x^2+5x\right)=0\\ 2\left(x+5\right)-\left[x\left(x+5\right)\right]=0\\ \left(2-x\right)\left(x+5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Answer 2

2(x + 5) - x^2 - 5x = 0

=> 2(x+5)-(2x+5x) = 0

=> 2(x+5) - x(x+5) = 0

=> (x+5).(2-x) = 0

=> x+5=0 => x = -5

     2-x=0 => x = 2

Vậy x=-5 và x=2