Question 55134
{{{x^2 - x = 12}}}
{{{x^2 - x - 12=0}}}
{{{x^2 - 4x + 3x - 4*3=0}}}
{{{x(x - 4) + 3(x - 4)=0}}}
{{{(x+3)(x - 4)=0}}}
Thus either (x + 3) = 0 i.e. x = -3 or x - 4 = 0 i.e. x = 4.