depends on wht the question is.usually, when you solve an equation for x, put the value back in and see if it satisfies the equation.for example.x^2+3x-4=0or, x^2 -x + 4x - 4 = 0 (3x = -x+4x)or, x(x-1) + 4(x-1)=0or, (x+4)(x-1) = 0thus, x = -4 or 1put this value back into the equation,when x = - 4,(-4)^2 + 3 times -4 - 4 = 16 -12-4=0 thus it satisfies the equation. the answer is correctagain, when x = 1,(1)^2 + 3 times 1 - 4=1+3-4=0thus, it satisfies the equation. the answer is correct.