50*x^2+20*x+1=0

Solution for 50*x^2+20*x+1=0 equation:

50x^2+20x+1=0

a = 50; b = 20; c = +1;
Δ = b2-4ac
Δ = 202-4·50·1
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{2}}{2*50}=\frac{-20-10\sqrt{2}}{100}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{2}}{2*50}=\frac{-20+10\sqrt{2}}{100}
$