The final squaring is beside the point. the only way (...)^2 would properly be 0 is for (...) to be 0. so sparkling up 5z^2 + 6z + 12 = 0 The quadratic formula z = (-b + or - sqrt(b^2 - 4ac))/2a with a = 5, b = 6, c = 12 will supply a destructive cost for b^2 - 4ac, meaning you get complicated roots. Do it your self, yet i think of you get z = -0.6 + 0.2*sqrt(fifty one)*i or z = -0.6 - 0.2*sqrt(fifty one)*i