5y(y+8)+2(y+1)=0

Solution for 5y(y+8)+2(y+1)=0 equation:

5y(y+8)+2(y+1)=0

We multiply parentheses

5y^2+40y+2y+2=0

We add all the numbers together, and all the variables

5y^2+42y+2=0

a = 5; b = 42; c = +2;
Δ = b2-4ac
Δ = 422-4·5·2
Δ = 1724
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:

$\sqrt{\Delta}=\sqrt{1724}=\sqrt{4*431}=\sqrt{4}*\sqrt{431}=2\sqrt{431}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{431}}{2*5}=\frac{-42-2\sqrt{431}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{431}}{2*5}=\frac{-42+2\sqrt{431}}{10}
$