x^2+20x+50=14

Solution for x^2+20x+50=14 equation:

x^2+20x+50=14

We move all terms to the left:

x^2+20x+50-(14)=0

We add all the numbers together, and all the variables

x^2+20x+36=0

a = 1; b = 20; c = +36;
Δ = b2-4ac
Δ = 202-4·1·36
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*1}=\frac{-36}{2}
=-18
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*1}=\frac{-4}{2}
=-2
$