Explanation:
if you did not make a mistake in the equation, then we have
-3x(x - 4) = -3x + 12
let's do the multiplication
-3x² + 12x = -3x + 12
and now let's combine all terms of the same type on one side of the equation to make it a "= 0" problem :
-3x² + 15x - 12 = 0
-x² + 5x - 4 = 0
this is a quadratic equation (because the highest exponent of the variable terms is "2" in x², so, things are squared or quadratic, hence the term "quadratic equation").
and it is one of the funny things in algebra :
am equation to the nth degree (that means the highest exponent of a variable term is n) has exactly n solutions (they might not be different, and they might not be members of R, but there are n solutions).
so, in our case, a quadratic equation has 2 solutions.
FYI - remember, the general solutions to such a quadratic equation
ax² + bx + c = 0
are
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = -1
b = 5
c = -4
x = (-5 ± sqrt(5² - 4×-1×-4))/(2×-1) =
= (-5 ± sqrt(25 - 16))/-2 = (-5 ± sqrt(9))/-2 =
= (-5 ± 3)/-2
x1 = (-5 + 3)/-2 = -2/-2 = 1
x2 = (-5 - 3)/-2 = -8/-2 = 4
so, our 2 solutions are x = 1 and x = 4.