14n-3n(4n+5)=-9+2(n-8)

Solution for 14n-3n(4n+5)=-9+2(n-8) equation:

14n-3n(4n+5)=-9+2(n-8)

We move all terms to the left:

14n-3n(4n+5)-(-9+2(n-8))=0

We multiply parentheses

-12n^2+14n-15n-(-9+2(n-8))=0


We calculate terms in parentheses: -(-9+2(n-8)), so:

-9+2(n-8)

determiningTheFunctionDomain
2(n-8)-9

We multiply parentheses

2n-16-9

We add all the numbers together, and all the variables

2n-25

Back to the equation:

-(2n-25)


We add all the numbers together, and all the variables

-12n^2-1n-(2n-25)=0

We get rid of parentheses

-12n^2-1n-2n+25=0

We add all the numbers together, and all the variables

-12n^2-3n+25=0

a = -12; b = -3; c = +25;
Δ = b2-4ac
Δ = -32-4·(-12)·25
Δ = 1209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1209}}{2*-12}=\frac{3-\sqrt{1209}}{-24}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1209}}{2*-12}=\frac{3+\sqrt{1209}}{-24}
$