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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1y^2+8y-2y+16)-8=0

We get rid of parentheses

-1y^2+8y-2y+16-8=0

We add all the numbers together, and all the variables

-1y^2+6y+8=0

a = -1; b = 6; c = +8;
Δ = b2-4ac
Δ = 62-4·(-1)·8
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*-1}=\frac{-6-2\sqrt{17}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*-1}=\frac{-6+2\sqrt{17}}{-2}
$