-6+19v+-3v2=0

Solution for -6+19v+-3v2=0 equation:

-6+19v+-3v^2=0

We add all the numbers together, and all the variables

-3v^2+19v=0

a = -3; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·(-3)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*-3}=\frac{-38}{-6}
=6+1/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*-3}=\frac{0}{-6}
=0
$