2v^2+19v-3=-4v^2-6

Solution for 2v^2+19v-3=-4v^2-6 equation:

2v^2+19v-3=-4v^2-6

We move all terms to the left:

2v^2+19v-3-(-4v^2-6)=0

We get rid of parentheses

2v^2+4v^2+19v+6-3=0

We add all the numbers together, and all the variables

6v^2+19v+3=0

a = 6; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·6·3
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*6}=\frac{-36}{12}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*6}=\frac{-2}{12}
=-1/6
$