11v^2-4v=16

Solution for 11v^2-4v=16 equation:

11v^2-4v=16

We move all terms to the left:

11v^2-4v-(16)=0

a = 11; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·11·(-16)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12\sqrt{5}}{2*11}=\frac{4-12\sqrt{5}}{22}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12\sqrt{5}}{2*11}=\frac{4+12\sqrt{5}}{22}
$