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

Solution for Y=-(x+2)(x-3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

-(-(+Y^2-3Y+2Y-6))=0


We calculate terms in parentheses: -(-(+Y^2-3Y+2Y-6)), so:

-(+Y^2-3Y+2Y-6)

We get rid of parentheses

-Y^2+3Y-2Y+6

We add all the numbers together, and all the variables

-1Y^2+Y+6

Back to the equation:

-(-1Y^2+Y+6)


We get rid of parentheses

1Y^2-Y-6=0

We add all the numbers together, and all the variables

Y^2-1Y-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$