14+3n=8n-3(n-4)n=1

Solution for 14+3n=8n-3(n-4)n=1 equation:

14+3n=8n-3(n-4)n=1

We move all terms to the left:

14+3n-(8n-3(n-4)n)=0


We calculate terms in parentheses: -(8n-3(n-4)n), so:

8n-3(n-4)n

We multiply parentheses

-3n^2+8n+12n

We add all the numbers together, and all the variables

-3n^2+20n

Back to the equation:

-(-3n^2+20n)


We get rid of parentheses

3n^2-20n+3n+14=0

We add all the numbers together, and all the variables

3n^2-17n+14=0

a = 3; b = -17; c = +14;
Δ = b2-4ac
Δ = -172-4·3·14
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-11}{2*3}=\frac{6}{6}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+11}{2*3}=\frac{28}{6}
=4+2/3
$