(1/6x+40)+(3x+26)=180

Solution for (1/6x+40)+(3x+26)=180 equation:

(1/6x+40)+(3x+26)=180

We move all terms to the left:

(1/6x+40)+(3x+26)-(180)=0


Domain of the equation: 6x+40)!=0

x∈R


We get rid of parentheses

1/6x+3x+40+26-180=0

We multiply all the terms by the denominator


3x*6x+40*6x+26*6x-180*6x+1=0

Wy multiply elements

18x^2+240x+156x-1080x+1=0

We add all the numbers together, and all the variables

18x^2-684x+1=0

a = 18; b = -684; c = +1;
Δ = b2-4ac
Δ = -6842-4·18·1
Δ = 467784
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:

$\sqrt{\Delta}=\sqrt{467784}=\sqrt{36*12994}=\sqrt{36}*\sqrt{12994}=6\sqrt{12994}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-684)-6\sqrt{12994}}{2*18}=\frac{684-6\sqrt{12994}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-684)+6\sqrt{12994}}{2*18}=\frac{684+6\sqrt{12994}}{36}
$