-2v^2+1v+6=0

Solution for -2v^2+1v+6=0 equation:

-2v^2+1v+6=0

We add all the numbers together, and all the variables

-2v^2+v+6=0

a = -2; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-2}=\frac{-8}{-4}
=+2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-2}=\frac{6}{-4}
=-1+1/2
$