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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-(5(+Y^2-4Y+3Y-12))=0


We calculate terms in parentheses: -(5(+Y^2-4Y+3Y-12)), so:

5(+Y^2-4Y+3Y-12)

We multiply parentheses

5Y^2-20Y+15Y-60

We add all the numbers together, and all the variables

5Y^2-5Y-60

Back to the equation:

-(5Y^2-5Y-60)


We get rid of parentheses

-5Y^2+5Y+60=0

a = -5; b = 5; c = +60;
Δ = b2-4ac
Δ = 52-4·(-5)·60
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-35}{2*-5}=\frac{-40}{-10}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+35}{2*-5}=\frac{30}{-10}
=-3
$