It's clear by looking at the constants that the constant in the expansion of the parentheses is -36 which will cancel out the +36. This also implies x=0 is a solution. If we expand the parentheses we get:
(x^2+2x-3)(x^2-8x+12)=x^4-8x^3+12x^2+2x^3-16x^2+24x-3x^2+24x-36=x^4-6x^3-7x^2+48x-36.
When we add 36 we get: x(x^3-6x^2-7x+48)=0 and x=0 is a solution as expected, since we lose the constant.
By inspection of the original equation, x=3 is also a solution. We can use synthetic division to find the quadratic:
3 | 1 -6 -7 48
1 3 -9 -48
1 -3 -16 | 0 = x^2-3x-16.
So we have the quadratic: x^2-3x-16=0; x^2-3x+9/4=16+9/4; (x-3/2)^2=73/4; x=3/2±√73/2=5.7720 and -2.7720 (approx).
So the complete solution is x=0, 3, 5.7720, -2.7720.