9v^2-4v-96=0

Solution for 9v^2-4v-96=0 equation:

9v^2-4v-96=0

a = 9; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·9·(-96)
Δ = 3472
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3472}=\sqrt{16*217}=\sqrt{16}*\sqrt{217}=4\sqrt{217}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{217}}{2*9}=\frac{4-4\sqrt{217}}{18}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{217}}{2*9}=\frac{4+4\sqrt{217}}{18}
$