33x^2-22x-96=0

Solution for 33x^2-22x-96=0 equation:

33x^2-22x-96=0

a = 33; b = -22; c = -96;
Δ = b2-4ac
Δ = -222-4·33·(-96)
Δ = 13156
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:

$\sqrt{\Delta}=\sqrt{13156}=\sqrt{4*3289}=\sqrt{4}*\sqrt{3289}=2\sqrt{3289}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{3289}}{2*33}=\frac{22-2\sqrt{3289}}{66}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{3289}}{2*33}=\frac{22+2\sqrt{3289}}{66}
$