Question 318135
(7z+5)^2+2(7z+5)-15=0
let (7z+5)be =x
x^2+2x-15=0
x^2+5x-3x-15=0
x(x+5)-3(x+5)=0
(x-3)(x+5)=0
replace the value of x
(7z+5-3)(7z+5+5)=0
(7z+2)(7z+10)=0
z=-2/7 or z=-10/7