2y(8y+3)=1

Solution for 2y(8y+3)=1 equation:

2y(8y+3)=1

We move all terms to the left:

2y(8y+3)-(1)=0

We multiply parentheses

16y^2+6y-1=0

a = 16; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·16·(-1)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*16}=\frac{-16}{32}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*16}=\frac{4}{32}
=1/8
$