Y=-5(x+4)(3x-3)

Solution for Y=-5(x+4)(3x-3) equation:

=-5(Y+4)(3Y-3)

We move all terms to the left:

-(-5(Y+4)(3Y-3))=0

We multiply parentheses ..

-(-5(+3Y^2-3Y+12Y-12))=0


We calculate terms in parentheses: -(-5(+3Y^2-3Y+12Y-12)), so:

-5(+3Y^2-3Y+12Y-12)

We multiply parentheses

-15Y^2+15Y-60Y+60

We add all the numbers together, and all the variables

-15Y^2-45Y+60

Back to the equation:

-(-15Y^2-45Y+60)


We get rid of parentheses

15Y^2+45Y-60=0

a = 15; b = 45; c = -60;
Δ = b2-4ac
Δ = 452-4·15·(-60)
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-75}{2*15}=\frac{-120}{30}
=-4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+75}{2*15}=\frac{30}{30}
=1
$