5-(n-4)=3n(n+2)

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

5-(n-4)=3n(n+2)

We move all terms to the left:

5-(n-4)-(3n(n+2))=0

We get rid of parentheses

-n-(3n(n+2))+4+5=0


We calculate terms in parentheses: -(3n(n+2)), so:

3n(n+2)

We multiply parentheses

3n^2+6n

Back to the equation:

-(3n^2+6n)


We add all the numbers together, and all the variables

-1n-(3n^2+6n)+9=0

We get rid of parentheses

-3n^2-1n-6n+9=0

We add all the numbers together, and all the variables

-3n^2-7n+9=0

a = -3; b = -7; c = +9;
Δ = b2-4ac
Δ = -72-4·(-3)·9
Δ = 157
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{-(-7)-\sqrt{157}}{2*-3}=\frac{7-\sqrt{157}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{157}}{2*-3}=\frac{7+\sqrt{157}}{-6}
$