5v2-4=19v

Solution for 5v2-4=19v equation:

5v^2-4=19v

We move all terms to the left:

5v^2-4-(19v)=0

a = 5; b = -19; c = -4;
Δ = b2-4ac
Δ = -192-4·5·(-4)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*5}=\frac{-2}{10}
=-1/5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*5}=\frac{40}{10}
=4
$