Y=(4n+5)(n-1)

Solution for Y=(4n+5)(n-1) equation:

=(4Y+5)(Y-1)

We move all terms to the left:

-((4Y+5)(Y-1))=0

We multiply parentheses ..

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


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

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

We get rid of parentheses

4Y^2-4Y+5Y-5

We add all the numbers together, and all the variables

4Y^2+Y-5

Back to the equation:

-(4Y^2+Y-5)


We get rid of parentheses

-4Y^2-Y+5=0

We add all the numbers together, and all the variables

-4Y^2-1Y+5=0

a = -4; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·(-4)·5
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{-(-1)-9}{2*-4}=\frac{-8}{-8}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*-4}=\frac{10}{-8}
=-1+1/4
$