0=3v^2+23v+30

Solution for 0=3v^2+23v+30 equation:

0=3v^2+23v+30

We move all terms to the left:

0-(3v^2+23v+30)=0

We add all the numbers together, and all the variables

-(3v^2+23v+30)=0

We get rid of parentheses

-3v^2-23v-30=0

a = -3; b = -23; c = -30;
Δ = b2-4ac
Δ = -232-4·(-3)·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-13}{2*-3}=\frac{10}{-6}
=-1+2/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+13}{2*-3}=\frac{36}{-6}
=-6
$