v^2-18v+9=0

Solution for v^2-18v+9=0 equation:

v^2-18v+9=0

a = 1; b = -18; c = +9;
Δ = b2-4ac
Δ = -182-4·1·9
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{2}}{2*1}=\frac{18-12\sqrt{2}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{2}}{2*1}=\frac{18+12\sqrt{2}}{2}
$