Y=(2-x)(15-4x)

Solution for Y=(2-x)(15-4x) equation:

=(2-Y)(15-4Y)

We move all terms to the left:

-((2-Y)(15-4Y))=0

We add all the numbers together, and all the variables

-((-1Y+2)(-4Y+15))=0

We multiply parentheses ..

-((+4Y^2-15Y-8Y+30))=0


We calculate terms in parentheses: -((+4Y^2-15Y-8Y+30)), so:

(+4Y^2-15Y-8Y+30)

We get rid of parentheses

4Y^2-15Y-8Y+30

We add all the numbers together, and all the variables

4Y^2-23Y+30

Back to the equation:

-(4Y^2-23Y+30)


We get rid of parentheses

-4Y^2+23Y-30=0

a = -4; b = 23; c = -30;
Δ = b2-4ac
Δ = 232-4·(-4)·(-30)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*-4}=\frac{-30}{-8}
=3+3/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*-4}=\frac{-16}{-8}
=+2
$