3x+1/4x+1=100

Solution for 3x+1/4x+1=100 equation:

3x+1/4x+1=100

We move all terms to the left:

3x+1/4x+1-(100)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We add all the numbers together, and all the variables

3x+1/4x-99=0

We multiply all the terms by the denominator


3x*4x-99*4x+1=0

Wy multiply elements

12x^2-396x+1=0

a = 12; b = -396; c = +1;
Δ = b2-4ac
Δ = -3962-4·12·1
Δ = 156768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156768}=\sqrt{16*9798}=\sqrt{16}*\sqrt{9798}=4\sqrt{9798}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-396)-4\sqrt{9798}}{2*12}=\frac{396-4\sqrt{9798}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-396)+4\sqrt{9798}}{2*12}=\frac{396+4\sqrt{9798}}{24}
$