-2v^2+13v+7=0

Solution for -2v^2+13v+7=0 equation:

-2v^2+13v+7=0

a = -2; b = 13; c = +7;
Δ = b2-4ac
Δ = 132-4·(-2)·7
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-15}{2*-2}=\frac{-28}{-4}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+15}{2*-2}=\frac{2}{-4}
=-1/2
$