v^2-16v+18=0

Solution for v^2-16v+18=0 equation:

v^2-16v+18=0

a = 1; b = -16; c = +18;
Δ = b2-4ac
Δ = -162-4·1·18
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{46}}{2*1}=\frac{16-2\sqrt{46}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{46}}{2*1}=\frac{16+2\sqrt{46}}{2}
$