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}
$