s^2+14s-31=3

Solution for s^2+14s-31=3 equation:

s^2+14s-31=3

We move all terms to the left:

s^2+14s-31-(3)=0

We add all the numbers together, and all the variables

s^2+14s-34=0

a = 1; b = 14; c = -34;
Δ = b2-4ac
Δ = 142-4·1·(-34)
Δ = 332
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{83}}{2*1}=\frac{-14-2\sqrt{83}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{83}}{2*1}=\frac{-14+2\sqrt{83}}{2}
$