Y=9(x-1)(2x+3)

Solution for Y=9(x-1)(2x+3) equation:

=9(Y-1)(2Y+3)

We move all terms to the left:

-(9(Y-1)(2Y+3))=0

We multiply parentheses ..

-(9(+2Y^2+3Y-2Y-3))=0


We calculate terms in parentheses: -(9(+2Y^2+3Y-2Y-3)), so:

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

We multiply parentheses

18Y^2+27Y-18Y-27

We add all the numbers together, and all the variables

18Y^2+9Y-27

Back to the equation:

-(18Y^2+9Y-27)


We get rid of parentheses

-18Y^2-9Y+27=0

a = -18; b = -9; c = +27;
Δ = b2-4ac
Δ = -92-4·(-18)·27
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-45}{2*-18}=\frac{-36}{-36}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+45}{2*-18}=\frac{54}{-36}
=-1+1/2
$