3v^2+4=4-2v

Solution for 3v^2+4=4-2v equation:

3v^2+4=4-2v

We move all terms to the left:

3v^2+4-(4-2v)=0

We add all the numbers together, and all the variables

3v^2-(-2v+4)+4=0

We get rid of parentheses

3v^2+2v-4+4=0

We add all the numbers together, and all the variables

3v^2+2v=0

a = 3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*3}=\frac{-4}{6}
=-2/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*3}=\frac{0}{6}
=0
$