(y+y)(y-6)=y^2+24

Solution for (y+y)(y-6)=y^2+24 equation:

(y+y)(y-6)=y^2+24

We move all terms to the left:

(y+y)(y-6)-(y^2+24)=0

We add all the numbers together, and all the variables

(+2y)(y-6)-(y^2+24)=0

We get rid of parentheses

-y^2+(+2y)(y-6)-24=0

We multiply parentheses ..

-y^2+(+2y^2-12y)-24=0

We add all the numbers together, and all the variables

-1y^2+(+2y^2-12y)-24=0

We get rid of parentheses

-1y^2+2y^2-12y-24=0

We add all the numbers together, and all the variables

y^2-12y-24=0

a = 1; b = -12; c = -24;
Δ = b2-4ac
Δ = -122-4·1·(-24)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{15}}{2*1}=\frac{12-4\sqrt{15}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{15}}{2*1}=\frac{12+4\sqrt{15}}{2}
$