x^2+20x+85=2

Solution for x^2+20x+85=2 equation:

x^2+20x+85=2

We move all terms to the left:

x^2+20x+85-(2)=0

We add all the numbers together, and all the variables

x^2+20x+83=0

a = 1; b = 20; c = +83;
Δ = b2-4ac
Δ = 202-4·1·83
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{17}}{2*1}=\frac{-20-2\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{17}}{2*1}=\frac{-20+2\sqrt{17}}{2}
$