4v^2=-20v

Solution for 4v^2=-20v equation:

4v^2=-20v

We move all terms to the left:

4v^2-(-20v)=0

We get rid of parentheses

4v^2+20v=0

a = 4; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·4·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*4}=\frac{-40}{8}
=-5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*4}=\frac{0}{8}
=0
$