-2v^2+5v+18=0

Solution for -2v^2+5v+18=0 equation:

-2v^2+5v+18=0

a = -2; b = 5; c = +18;
Δ = b2-4ac
Δ = 52-4·(-2)·18
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*-2}=\frac{-18}{-4}
=4+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-2}=\frac{8}{-4}
=-2
$