1/4x-8=2+x

Solution for 1/4x-8=2+x equation:

1/4x-8=2+x

We move all terms to the left:

1/4x-8-(2+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+2)-8=0

We get rid of parentheses

1/4x-x-2-8=0

We multiply all the terms by the denominator


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

Wy multiply elements

-4x^2-8x-32x+1=0

We add all the numbers together, and all the variables

-4x^2-40x+1=0

a = -4; b = -40; c = +1;
Δ = b2-4ac
Δ = -402-4·(-4)·1
Δ = 1616
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:

$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{101}}{2*-4}=\frac{40-4\sqrt{101}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{101}}{2*-4}=\frac{40+4\sqrt{101}}{-8}
$