16=4v^2

Solution for 16=4v^2 equation:

16=4v^2

We move all terms to the left:

16-(4v^2)=0

a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-4}=\frac{-16}{-8}
=+2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8}
=-2
$