8y(y+2)=1

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

8y(y+2)=1

We move all terms to the left:

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

We multiply parentheses

8y^2+16y-1=0

a = 8; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·8·(-1)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12\sqrt{2}}{2*8}=\frac{-16-12\sqrt{2}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{2}}{2*8}=\frac{-16+12\sqrt{2}}{16}
$