-3x^2+408x=2232

Solution for -3x^2+408x=2232 equation:

-3x^2+408x=2232

We move all terms to the left:

-3x^2+408x-(2232)=0

a = -3; b = 408; c = -2232;
Δ = b2-4ac
Δ = 4082-4·(-3)·(-2232)
Δ = 139680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{139680}=\sqrt{144*970}=\sqrt{144}*\sqrt{970}=12\sqrt{970}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(408)-12\sqrt{970}}{2*-3}=\frac{-408-12\sqrt{970}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(408)+12\sqrt{970}}{2*-3}=\frac{-408+12\sqrt{970}}{-6}
$