v^2+18v+53=0

Solution for v^2+18v+53=0 equation:

v^2+18v+53=0

a = 1; b = 18; c = +53;
Δ = b2-4ac
Δ = 182-4·1·53
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{7}}{2*1}=\frac{-18-4\sqrt{7}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{7}}{2*1}=\frac{-18+4\sqrt{7}}{2}
$