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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1v^2-7v+12=0

a = -1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·(-1)·12
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{97}}{2*-1}=\frac{7-\sqrt{97}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{97}}{2*-1}=\frac{7+\sqrt{97}}{-2}
$