1/4x=3+x

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

1/4x=3+x

We move all terms to the left:

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


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/4x-x-3=0

We multiply all the terms by the denominator


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

Wy multiply elements

-4x^2-12x+1=0

a = -4; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·(-4)·1
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*-4}=\frac{12-4\sqrt{10}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*-4}=\frac{12+4\sqrt{10}}{-8}
$