-3x^2+102x-321=0

Solution for -3x^2+102x-321=0 equation:

-3x^2+102x-321=0

a = -3; b = 102; c = -321;
Δ = b2-4ac
Δ = 1022-4·(-3)·(-321)
Δ = 6552
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:

$\sqrt{\Delta}=\sqrt{6552}=\sqrt{36*182}=\sqrt{36}*\sqrt{182}=6\sqrt{182}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-6\sqrt{182}}{2*-3}=\frac{-102-6\sqrt{182}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+6\sqrt{182}}{2*-3}=\frac{-102+6\sqrt{182}}{-6}
$