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

Solution for -2v2-v+12=-3v2+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-(-3v^2+6v)-1v+12=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

v^2-7v+12=0

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

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*1}=\frac{6}{2}
=3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$