19x2-20x+5=0

Solution for 19x2-20x+5=0 equation:

19x^2-20x+5=0

a = 19; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·19·5
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{5}}{2*19}=\frac{20-2\sqrt{5}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{5}}{2*19}=\frac{20+2\sqrt{5}}{38}
$