(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+5)!=0

x∈R



Domain of the equation: 4x+2)!=0

x∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


x*4x+x*4x-4*4x+1*4x+5*4x-2*4x+1-9=0

We add all the numbers together, and all the variables

x*4x+x*4x-4*4x+1*4x+5*4x-2*4x-8=0

Wy multiply elements

4x^2+4x^2-16x+4x+20x-8x-8=0

We add all the numbers together, and all the variables

8x^2-8=0

a = 8; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·8·(-8)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*8}=\frac{-16}{16}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*8}=\frac{16}{16}
=1
$