8=-18+(3/8)(16-40n)

Solution for 8=-18+(3/8)(16-40n) equation:

8=-18+(3/8)(16-40n)

We move all terms to the left:

8-(-18+(3/8)(16-40n))=0


Domain of the equation: 8)(16-40n))!=0

n∈R


We add all the numbers together, and all the variables

-(-18+(+3/8)(-40n+16))+8=0

We multiply parentheses ..

-(-18+(-120n^2+3/8*16))+8=0

We multiply all the terms by the denominator


-(-18+(-120n^2+3+8*8*16))=0


We calculate terms in parentheses: -(-18+(-120n^2+3+8*8*16)), so:

-18+(-120n^2+3+8*8*16)

determiningTheFunctionDomain
(-120n^2+3+8*8*16)-18

We get rid of parentheses

-120n^2+3-18+8*8*16

We add all the numbers together, and all the variables

-120n^2+1009

Back to the equation:

-(-120n^2+1009)


We get rid of parentheses

120n^2-1009=0

a = 120; b = 0; c = -1009;
Δ = b2-4ac
Δ = 02-4·120·(-1009)
Δ = 484320
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:

$\sqrt{\Delta}=\sqrt{484320}=\sqrt{16*30270}=\sqrt{16}*\sqrt{30270}=4\sqrt{30270}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30270}}{2*120}=\frac{0-4\sqrt{30270}}{240}
=-\frac{4\sqrt{30270}}{240}
=-\frac{\sqrt{30270}}{60}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30270}}{2*120}=\frac{0+4\sqrt{30270}}{240}
=\frac{4\sqrt{30270}}{240}
=\frac{\sqrt{30270}}{60}
$