1/312x+18=4x-20

Solution for 1/312x+18=4x-20 equation:

1/312x+18=4x-20

We move all terms to the left:

1/312x+18-(4x-20)=0


Domain of the equation: 312x!=0

x!=0/312

x!=0

x∈R


We get rid of parentheses

1/312x-4x+20+18=0

We multiply all the terms by the denominator


-4x*312x+20*312x+18*312x+1=0

Wy multiply elements

-1248x^2+6240x+5616x+1=0

We add all the numbers together, and all the variables

-1248x^2+11856x+1=0

a = -1248; b = 11856; c = +1;
Δ = b2-4ac
Δ = 118562-4·(-1248)·1
Δ = 140569728
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:

$\sqrt{\Delta}=\sqrt{140569728}=\sqrt{64*2196402}=\sqrt{64}*\sqrt{2196402}=8\sqrt{2196402}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11856)-8\sqrt{2196402}}{2*-1248}=\frac{-11856-8\sqrt{2196402}}{-2496}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11856)+8\sqrt{2196402}}{2*-1248}=\frac{-11856+8\sqrt{2196402}}{-2496}
$