-2v^2+4v+45=0

Solution for -2v^2+4v+45=0 equation:

-2v^2+4v+45=0

a = -2; b = 4; c = +45;
Δ = b2-4ac
Δ = 42-4·(-2)·45
Δ = 376
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:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{94}}{2*-2}=\frac{-4-2\sqrt{94}}{-4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{94}}{2*-2}=\frac{-4+2\sqrt{94}}{-4}
$