3r2-6r-93=0

Solution for 3r2-6r-93=0 equation:

3r^2-6r-93=0

a = 3; b = -6; c = -93;
Δ = b2-4ac
Δ = -62-4·3·(-93)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-24\sqrt{2}}{2*3}=\frac{6-24\sqrt{2}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+24\sqrt{2}}{2*3}=\frac{6+24\sqrt{2}}{6}
$