1/3x+4=20x=48

Solution for 1/3x+4=20x=48 equation:

1/3x+4=20x=48

We move all terms to the left:

1/3x+4-(20x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

-20x+1/3x+4=0

We multiply all the terms by the denominator


-20x*3x+4*3x+1=0

Wy multiply elements

-60x^2+12x+1=0

a = -60; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-60)·1
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*-60}=\frac{-12-8\sqrt{6}}{-120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*-60}=\frac{-12+8\sqrt{6}}{-120}
$