1/4n+12=1/2n-3

Solution for 1/4n+12=1/2n-3 equation:

1/4n+12=1/2n-3

We move all terms to the left:

1/4n+12-(1/2n-3)=0


Domain of the equation: 4n!=0

n!=0/4

n!=0

n∈R



Domain of the equation: 2n-3)!=0

n∈R


We get rid of parentheses

1/4n-1/2n+3+12=0

We calculate fractions


2n/8n^2+(-4n)/8n^2+3+12=0

We add all the numbers together, and all the variables

2n/8n^2+(-4n)/8n^2+15=0

We multiply all the terms by the denominator


2n+(-4n)+15*8n^2=0

Wy multiply elements

120n^2+2n+(-4n)=0

We get rid of parentheses

120n^2+2n-4n=0

We add all the numbers together, and all the variables

120n^2-2n=0

a = 120; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·120·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*120}=\frac{0}{240}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*120}=\frac{4}{240}
=1/60
$