-2v-v+12=3v2+6v

Solution for -2v-v+12=3v2+6v equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3v^2-9v+12=0

a = -3; b = -9; c = +12;
Δ = b2-4ac
Δ = -92-4·(-3)·12
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*-3}=\frac{-6}{-6}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*-3}=\frac{24}{-6}
=-4
$