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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-(2(+Y^2-5Y-3Y+15))=0


We calculate terms in parentheses: -(2(+Y^2-5Y-3Y+15)), so:

2(+Y^2-5Y-3Y+15)

We multiply parentheses

2Y^2-10Y-6Y+30

We add all the numbers together, and all the variables

2Y^2-16Y+30

Back to the equation:

-(2Y^2-16Y+30)


We get rid of parentheses

-2Y^2+16Y-30=0

a = -2; b = 16; c = -30;
Δ = b2-4ac
Δ = 162-4·(-2)·(-30)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*-2}=\frac{-20}{-4}
=+5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*-2}=\frac{-12}{-4}
=+3
$