4v^2+8v-96=3

Solution for 4v^2+8v-96=3 equation:

4v^2+8v-96=3

We move all terms to the left:

4v^2+8v-96-(3)=0

We add all the numbers together, and all the variables

4v^2+8v-99=0

a = 4; b = 8; c = -99;
Δ = b2-4ac
Δ = 82-4·4·(-99)
Δ = 1648
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:

$\sqrt{\Delta}=\sqrt{1648}=\sqrt{16*103}=\sqrt{16}*\sqrt{103}=4\sqrt{103}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{103}}{2*4}=\frac{-8-4\sqrt{103}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{103}}{2*4}=\frac{-8+4\sqrt{103}}{8}
$