Y=(3x+12)(3x+10)

Solution for Y=(3x+12)(3x+10) equation:

=(3Y+12)(3Y+10)

We move all terms to the left:

-((3Y+12)(3Y+10))=0

We multiply parentheses ..

-((+9Y^2+30Y+36Y+120))=0


We calculate terms in parentheses: -((+9Y^2+30Y+36Y+120)), so:

(+9Y^2+30Y+36Y+120)

We get rid of parentheses

9Y^2+30Y+36Y+120

We add all the numbers together, and all the variables

9Y^2+66Y+120

Back to the equation:

-(9Y^2+66Y+120)


We get rid of parentheses

-9Y^2-66Y-120=0

a = -9; b = -66; c = -120;
Δ = b2-4ac
Δ = -662-4·(-9)·(-120)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-66)-6}{2*-9}=\frac{60}{-18}
=-3+1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-66)+6}{2*-9}=\frac{72}{-18}
=-4
$