V=3.13(4a)(3a+2)

Solution for V=3.13(4a)(3a+2) equation:

=3.13(4V)(3V+2)

We move all terms to the left:

-(3.13(4V)(3V+2))=0


We calculate terms in parentheses: -(3.134V(3V+2)), so:

3.134V(3V+2)

We multiply parentheses

9V^2+6V

Back to the equation:

-(9V^2+6V)


We get rid of parentheses

-9V^2-6V=0

a = -9; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-9)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-9}=\frac{0}{-18}
=0
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-9}=\frac{12}{-18}
=-2/3
$