4y^2-20y+19=0

Solution for 4y^2-20y+19=0 equation:

4y^2-20y+19=0

a = 4; b = -20; c = +19;
Δ = b2-4ac
Δ = -202-4·4·19
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{6}}{2*4}=\frac{20-4\sqrt{6}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{6}}{2*4}=\frac{20+4\sqrt{6}}{8}
$