(y+2)(y-4)=6

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

(y+2)(y-4)=6

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

y^2-4y+2y-8-6=0

We add all the numbers together, and all the variables

y^2-2y-14=0

a = 1; b = -2; c = -14;
Δ = b2-4ac
Δ = -22-4·1·(-14)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{15}}{2*1}=\frac{2-2\sqrt{15}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{15}}{2*1}=\frac{2+2\sqrt{15}}{2}
$