v^2+18v+45=0

Solution for v^2+18v+45=0 equation:

v^2+18v+45=0

a = 1; b = 18; c = +45;
Δ = b2-4ac
Δ = 182-4·1·45
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12}{2*1}=\frac{-30}{2}
=-15
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*1}=\frac{-6}{2}
=-3
$