16y^2-80y+19=0

Solution for 16y^2-80y+19=0 equation:

16y^2-80y+19=0

a = 16; b = -80; c = +19;
Δ = b2-4ac
Δ = -802-4·16·19
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-72}{2*16}=\frac{8}{32}
=1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+72}{2*16}=\frac{152}{32}
=4+3/4
$