-2v^2-v+36=0

Solution for -2v^2-v+36=0 equation:

-2v^2-v+36=0

We add all the numbers together, and all the variables

-2v^2-1v+36=0

a = -2; b = -1; c = +36;
Δ = b2-4ac
Δ = -12-4·(-2)·36
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*-2}=\frac{-16}{-4}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*-2}=\frac{18}{-4}
=-4+1/2
$