64+16v=v^2

Solution for 64+16v=v^2 equation:

64+16v=v^2

We move all terms to the left:

64+16v-(v^2)=0

determiningTheFunctionDomain
-v^2+16v+64=0

We add all the numbers together, and all the variables

-1v^2+16v+64=0

a = -1; b = 16; c = +64;
Δ = b2-4ac
Δ = 162-4·(-1)·64
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*-1}=\frac{-16-16\sqrt{2}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*-1}=\frac{-16+16\sqrt{2}}{-2}
$