-2v^2+13v+25=0

Solution for -2v^2+13v+25=0 equation:

-2v^2+13v+25=0

a = -2; b = 13; c = +25;
Δ = b2-4ac
Δ = 132-4·(-2)·25
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3\sqrt{41}}{2*-2}=\frac{-13-3\sqrt{41}}{-4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3\sqrt{41}}{2*-2}=\frac{-13+3\sqrt{41}}{-4}
$