v^2-2v+16=32

Solution for v^2-2v+16=32 equation:

v^2-2v+16=32

We move all terms to the left:

v^2-2v+16-(32)=0

We add all the numbers together, and all the variables

v^2-2v-16=0

a = 1; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·1·(-16)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*1}=\frac{2-2\sqrt{17}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*1}=\frac{2+2\sqrt{17}}{2}
$