Question 537004
x+1=sqrt7x-5

square both sides of the equation

(x+1)^2=(sqrt7x-5)^2

x^2+2x+1 = 7x-5

x^2-5x+6=0
x^2-3x-2x+6=0
x(x-3)-2(x-3)=0
(x-3)(x-2)=0
x=3, or 2