V=3b+2b2

Solution for V=3b+2b2 equation:

=3V+2V^2

We move all terms to the left:

-(3V+2V^2)=0

We get rid of parentheses

-2V^2-3V=0

a = -2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-2)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-2}=\frac{0}{-4}
=0
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-2}=\frac{6}{-4}
=-1+1/2
$