(y-2)(y+4)=3

Solution for (y-2)(y+4)=3 equation:

(y-2)(y+4)=3

We move all terms to the left:

(y-2)(y+4)-(3)=0

We multiply parentheses ..

(+y^2+4y-2y-8)-3=0

We get rid of parentheses

y^2+4y-2y-8-3=0

We add all the numbers together, and all the variables

y^2+2y-11=0

a = 1; b = 2; c = -11;
Δ = b2-4ac
Δ = 22-4·1·(-11)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{3}}{2*1}=\frac{-2-4\sqrt{3}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{3}}{2*1}=\frac{-2+4\sqrt{3}}{2}
$