9(n-4)7n=32-2(n+8)

Solution for 9(n-4)7n=32-2(n+8) equation:

9(n-4)7n=32-2(n+8)

We move all terms to the left:

9(n-4)7n-(32-2(n+8))=0

We multiply parentheses

63n^2-252n-(32-2(n+8))=0


We calculate terms in parentheses: -(32-2(n+8)), so:

32-2(n+8)

determiningTheFunctionDomain
-2(n+8)+32

We multiply parentheses

-2n-16+32

We add all the numbers together, and all the variables

-2n+16

Back to the equation:

-(-2n+16)


We get rid of parentheses

63n^2-252n+2n-16=0

We add all the numbers together, and all the variables

63n^2-250n-16=0

a = 63; b = -250; c = -16;
Δ = b2-4ac
Δ = -2502-4·63·(-16)
Δ = 66532
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{66532}=\sqrt{4*16633}=\sqrt{4}*\sqrt{16633}=2\sqrt{16633}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-250)-2\sqrt{16633}}{2*63}=\frac{250-2\sqrt{16633}}{126}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-250)+2\sqrt{16633}}{2*63}=\frac{250+2\sqrt{16633}}{126}
$