-4+100v^2=21

Solution for -4+100v^2=21 equation:

-4+100v^2=21

We move all terms to the left:

-4+100v^2-(21)=0

We add all the numbers together, and all the variables

100v^2-25=0

a = 100; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·100·(-25)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-100}{2*100}=\frac{-100}{200}
=-1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+100}{2*100}=\frac{100}{200}
=1/2
$