8y+8y(10y+5)=80

Solution for 8y+8y(10y+5)=80 equation:

8y+8y(10y+5)=80

We move all terms to the left:

8y+8y(10y+5)-(80)=0

We multiply parentheses

80y^2+8y+40y-80=0

We add all the numbers together, and all the variables

80y^2+48y-80=0

a = 80; b = 48; c = -80;
Δ = b2-4ac
Δ = 482-4·80·(-80)
Δ = 27904
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:

$\sqrt{\Delta}=\sqrt{27904}=\sqrt{256*109}=\sqrt{256}*\sqrt{109}=16\sqrt{109}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{109}}{2*80}=\frac{-48-16\sqrt{109}}{160}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{109}}{2*80}=\frac{-48+16\sqrt{109}}{160}
$