-2v^2-10v+40=0

Solution for -2v^2-10v+40=0 equation:

-2v^2-10v+40=0

a = -2; b = -10; c = +40;
Δ = b2-4ac
Δ = -102-4·(-2)·40
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{105}}{2*-2}=\frac{10-2\sqrt{105}}{-4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{105}}{2*-2}=\frac{10+2\sqrt{105}}{-4}
$