16+3v2=19v

Solution for 16+3v2=19v equation:

16+3v^2=19v

We move all terms to the left:

16+3v^2-(19v)=0

a = 3; b = -19; c = +16;
Δ = b2-4ac
Δ = -192-4·3·16
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*3}=\frac{6}{6}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*3}=\frac{32}{6}
=5+1/3
$