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 = 14256
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:

$\sqrt{\Delta}=\sqrt{14256}=\sqrt{1296*11}=\sqrt{1296}*\sqrt{11}=36\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-36\sqrt{11}}{2*3}=\frac{-102-36\sqrt{11}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+36\sqrt{11}}{2*3}=\frac{-102+36\sqrt{11}}{6}
$