3(n/4)*3n=42

Solution for 3(n/4)*3n=42 equation:

3(n/4)*3n=42

We move all terms to the left:

3(n/4)*3n-(42)=0


Domain of the equation: 4)*3n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

3(+n/4)*3n-42=0

We multiply parentheses

9n^2-42=0

a = 9; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·9·(-42)
Δ = 1512
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:

$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{42}}{2*9}=\frac{0-6\sqrt{42}}{18}
=-\frac{6\sqrt{42}}{18}
=-\frac{\sqrt{42}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{42}}{2*9}=\frac{0+6\sqrt{42}}{18}
=\frac{6\sqrt{42}}{18}
=\frac{\sqrt{42}}{3}
$