Y=(x-3)(x+12)

Solution for Y=(x-3)(x+12) equation:

=(Y-3)(Y+12)

We move all terms to the left:

-((Y-3)(Y+12))=0

We multiply parentheses ..

-((+Y^2+12Y-3Y-36))=0


We calculate terms in parentheses: -((+Y^2+12Y-3Y-36)), so:

(+Y^2+12Y-3Y-36)

We get rid of parentheses

Y^2+12Y-3Y-36

We add all the numbers together, and all the variables

Y^2+9Y-36

Back to the equation:

-(Y^2+9Y-36)


We get rid of parentheses

-Y^2-9Y+36=0

We add all the numbers together, and all the variables

-1Y^2-9Y+36=0

a = -1; b = -9; c = +36;
Δ = b2-4ac
Δ = -92-4·(-1)·36
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*-1}=\frac{-6}{-2}
=+3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*-1}=\frac{24}{-2}
=-12
$