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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We multiply parentheses

3Y^2+6Y+15Y+30

We add all the numbers together, and all the variables

3Y^2+21Y+30

Back to the equation:

-(3Y^2+21Y+30)


We get rid of parentheses

-3Y^2-21Y-30=0

a = -3; b = -21; c = -30;
Δ = b2-4ac
Δ = -212-4·(-3)·(-30)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-9}{2*-3}=\frac{12}{-6}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+9}{2*-3}=\frac{30}{-6}
=-5
$