v^2+14v+31=0

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Solution for v^2+14v+31=0 equation:



v^2+14v+31=0
a = 1; b = 14; c = +31;
Δ = b2-4ac
Δ = 142-4·1·31
Δ = 72
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:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6\sqrt{2}}{2*1}=\frac{-14-6\sqrt{2}}{2} $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6\sqrt{2}}{2*1}=\frac{-14+6\sqrt{2}}{2} $

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