3(n-8)-5=10-4/7n

Solution for 3(n-8)-5=10-4/7n equation:

3(n-8)-5=10-4/7n

We move all terms to the left:

3(n-8)-5-(10-4/7n)=0


Domain of the equation: 7n)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

3(n-8)-(-4/7n+10)-5=0

We multiply parentheses

3n-(-4/7n+10)-24-5=0

We get rid of parentheses

3n+4/7n-10-24-5=0

We multiply all the terms by the denominator


3n*7n-10*7n-24*7n-5*7n+4=0

Wy multiply elements

21n^2-70n-168n-35n+4=0

We add all the numbers together, and all the variables

21n^2-273n+4=0

a = 21; b = -273; c = +4;
Δ = b2-4ac
Δ = -2732-4·21·4
Δ = 74193
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{-(-273)-\sqrt{74193}}{2*21}=\frac{273-\sqrt{74193}}{42}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-273)+\sqrt{74193}}{2*21}=\frac{273+\sqrt{74193}}{42}
$