x^2+20x+100=14

Solution for x^2+20x+100=14 equation:

x^2+20x+100=14

We move all terms to the left:

x^2+20x+100-(14)=0

We add all the numbers together, and all the variables

x^2+20x+86=0

a = 1; b = 20; c = +86;
Δ = b2-4ac
Δ = 202-4·1·86
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{14}}{2*1}=\frac{-20-2\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{14}}{2*1}=\frac{-20+2\sqrt{14}}{2}
$