v^2=9v+65

Solution for v^2=9v+65 equation:

v^2=9v+65

We move all terms to the left:

v^2-(9v+65)=0

We get rid of parentheses

v^2-9v-65=0

a = 1; b = -9; c = -65;
Δ = b2-4ac
Δ = -92-4·1·(-65)
Δ = 341
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{341}}{2*1}=\frac{9-\sqrt{341}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{341}}{2*1}=\frac{9+\sqrt{341}}{2}
$