2v^2=16v+91

Solution for 2v^2=16v+91 equation:

2v^2=16v+91

We move all terms to the left:

2v^2-(16v+91)=0

We get rid of parentheses

2v^2-16v-91=0

a = 2; b = -16; c = -91;
Δ = b2-4ac
Δ = -162-4·2·(-91)
Δ = 984
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:

$\sqrt{\Delta}=\sqrt{984}=\sqrt{4*246}=\sqrt{4}*\sqrt{246}=2\sqrt{246}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{246}}{2*2}=\frac{16-2\sqrt{246}}{4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{246}}{2*2}=\frac{16+2\sqrt{246}}{4}
$