v^2=-18+9v

Solution for v^2=-18+9v equation:

v^2=-18+9v

We move all terms to the left:

v^2-(-18+9v)=0

We add all the numbers together, and all the variables

v^2-(9v-18)=0

We get rid of parentheses

v^2-9v+18=0

a = 1; b = -9; c = +18;
Δ = b2-4ac
Δ = -92-4·1·18
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{-(-9)-3}{2*1}=\frac{6}{2}
=3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*1}=\frac{12}{2}
=6
$