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

Solution for -2v^2+4v+19=0 equation:

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

a = -2; b = 4; c = +19;
Δ = b2-4ac
Δ = 42-4·(-2)·19
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{42}}{2*-2}=\frac{-4-2\sqrt{42}}{-4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{42}}{2*-2}=\frac{-4+2\sqrt{42}}{-4}
$