-2v^2-v+12=-3v+6

Solution for -2v^2-v+12=-3v+6 equation:

-2v^2-v+12=-3v+6

We move all terms to the left:

-2v^2-v+12-(-3v+6)=0

We add all the numbers together, and all the variables

-2v^2-1v-(-3v+6)+12=0

We get rid of parentheses

-2v^2-1v+3v-6+12=0

We add all the numbers together, and all the variables

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

a = -2; b = 2; c = +6;
Δ = b2-4ac
Δ = 22-4·(-2)·6
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*-2}=\frac{-2-2\sqrt{13}}{-4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*-2}=\frac{-2+2\sqrt{13}}{-4}
$