Y=3(3x+5)(x-2)

Solution for Y=3(3x+5)(x-2) equation:

=3(3Y+5)(Y-2)

We move all terms to the left:

-(3(3Y+5)(Y-2))=0

We multiply parentheses ..

-(3(+3Y^2-6Y+5Y-10))=0


We calculate terms in parentheses: -(3(+3Y^2-6Y+5Y-10)), so:

3(+3Y^2-6Y+5Y-10)

We multiply parentheses

9Y^2-18Y+15Y-30

We add all the numbers together, and all the variables

9Y^2-3Y-30

Back to the equation:

-(9Y^2-3Y-30)


We get rid of parentheses

-9Y^2+3Y+30=0

a = -9; b = 3; c = +30;
Δ = b2-4ac
Δ = 32-4·(-9)·30
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-33}{2*-9}=\frac{-36}{-18}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+33}{2*-9}=\frac{30}{-18}
=-1+2/3
$