(8y+2)4y+6=68

Solution for (8y+2)4y+6=68 equation:

(8y+2)4y+6=68

We move all terms to the left:

(8y+2)4y+6-(68)=0

We add all the numbers together, and all the variables

(8y+2)4y-62=0

We multiply parentheses

32y^2+8y-62=0

a = 32; b = 8; c = -62;
Δ = b2-4ac
Δ = 82-4·32·(-62)
Δ = 8000
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:

$\sqrt{\Delta}=\sqrt{8000}=\sqrt{1600*5}=\sqrt{1600}*\sqrt{5}=40\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40\sqrt{5}}{2*32}=\frac{-8-40\sqrt{5}}{64}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40\sqrt{5}}{2*32}=\frac{-8+40\sqrt{5}}{64}
$