Use quadratic formula to solve:

3y + $\dfrac{5}{16y}$ = 2

Given,

3y + $\dfrac{5}{16y}$ = 2

$\dfrac{48y^2 + 5}{16y}$ = 2

48y² + 5 = 32y

48y² - 32y + 5 = 0

Comparing 48y² - 32y + 5 = 0 with ax² + bx + c = 0 we get,

a = 48, b = -32 and c = 5.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-32) \pm \sqrt{(-32)^2 - 4(48)(5)}}{2(48)} \\[1em] = \dfrac{32 \pm \sqrt{1024 - 960}}{96} \\[1em] = \dfrac{32 \pm \sqrt{64}}{96} \\[1em] = \dfrac{32 \pm 8}{96} \\[1em] = \dfrac{32 + 8}{96} \text{ or } \dfrac{32 - 8}{96} \\[1em] = \dfrac{40}{96} \text{ or } \dfrac{24}{96} \\[1em] = \dfrac{5}{12} \text{ or } \dfrac{1}{4}.$

Hence, x = $\dfrac{5}{12} \text{ or } \dfrac{1}{4}$.