b^2+20b+50=0

Solution for b^2+20b+50=0 equation:

b^2+20b+50=0

a = 1; b = 20; c = +50;
Δ = b2-4ac
Δ = 202-4·1·50
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{2}}{2*1}=\frac{-20-10\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{2}}{2*1}=\frac{-20+10\sqrt{2}}{2}
$