-3x^2+24x=36

Solution for -3x^2+24x=36 equation:

-3x^2+24x=36

We move all terms to the left:

-3x^2+24x-(36)=0

a = -3; b = 24; c = -36;
Δ = b2-4ac
Δ = 242-4·(-3)·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12}{2*-3}=\frac{-12}{-6}
=+2
$