a^2+2a+20=180

Solution for a^2+2a+20=180 equation:

a^2+2a+20=180

We move all terms to the left:

a^2+2a+20-(180)=0

We add all the numbers together, and all the variables

a^2+2a-160=0

a = 1; b = 2; c = -160;
Δ = b2-4ac
Δ = 22-4·1·(-160)
Δ = 644
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{161}}{2*1}=\frac{-2-2\sqrt{161}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{161}}{2*1}=\frac{-2+2\sqrt{161}}{2}
$