16y^2-18y+3=0

Solution for 16y^2-18y+3=0 equation:

16y^2-18y+3=0

a = 16; b = -18; c = +3;
Δ = b2-4ac
Δ = -182-4·16·3
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{33}}{2*16}=\frac{18-2\sqrt{33}}{32}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{33}}{2*16}=\frac{18+2\sqrt{33}}{32}
$