4v^2=-28v

Solution for 4v^2=-28v equation:

4v^2=-28v

We move all terms to the left:

4v^2-(-28v)=0

We get rid of parentheses

4v^2+28v=0

a = 4; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·4·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*4}=\frac{-56}{8}
=-7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*4}=\frac{0}{8}
=0
$