-3=-v^2+9v

Solution for -3=-v^2+9v equation:

-3=-v^2+9v

We move all terms to the left:

-3-(-v^2+9v)=0

We get rid of parentheses

v^2-9v-3=0

a = 1; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·1·(-3)
Δ = 93
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{93}}{2*1}=\frac{9-\sqrt{93}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{93}}{2*1}=\frac{9+\sqrt{93}}{2}
$