16v^2=1

Solution for 16v^2=1 equation:

16v^2=1

We move all terms to the left:

16v^2-(1)=0

a = 16; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·16·(-1)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*16}=\frac{-8}{32}
=-1/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*16}=\frac{8}{32}
=1/4
$