(x-4)+(x+1)+1/4x+5)=9/4x+2

Solution for (x-4)+(x+1)+1/4x+5)=9/4x+2 equation:

(x-4)+(x+1)+1/4x+5)=9/4x+2

We move all terms to the left:

(x-4)+(x+1)+1/4x+5)-(9/4x+2)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We add all the numbers together, and all the variables

(x-4)+(x+1)+1/4x+5)-(9/4x=0

We get rid of parentheses

x+x+1/4x+5)-(9/4x-4+1=0

We multiply all the terms by the denominator


x*4x+x*4x-4*4x+1*4x+1+5)-(9=0

We add all the numbers together, and all the variables

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

Wy multiply elements

4x^2+4x^2-16x+4x=0

We add all the numbers together, and all the variables

8x^2-12x=0

a = 8; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·8·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*8}=\frac{0}{16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*8}=\frac{24}{16}
=1+1/2
$