v^2+10=9v

Solution for v^2+10=9v equation:

v^2+10=9v

We move all terms to the left:

v^2+10-(9v)=0

a = 1; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·1·10
Δ = 41
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{41}}{2*1}=\frac{9-\sqrt{41}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41}}{2*1}=\frac{9+\sqrt{41}}{2}
$