Y2-3=(y-2)(y+4)

Solution for Y2-3=(y-2)(y+4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-((Y-2)(Y+4))-1=0

We multiply parentheses ..

-((+Y^2+4Y-2Y-8))-1=0


We calculate terms in parentheses: -((+Y^2+4Y-2Y-8)), so:

(+Y^2+4Y-2Y-8)

We get rid of parentheses

Y^2+4Y-2Y-8

We add all the numbers together, and all the variables

Y^2+2Y-8

Back to the equation:

-(Y^2+2Y-8)


We get rid of parentheses

-Y^2-2Y+8-1=0

We add all the numbers together, and all the variables

-1Y^2-2Y+7=0

a = -1; b = -2; c = +7;
Δ = b2-4ac
Δ = -22-4·(-1)·7
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{2}}{2*-1}=\frac{2-4\sqrt{2}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{2}}{2*-1}=\frac{2+4\sqrt{2}}{-2}
$