(x-10)+(2x-123)+(1/2x+5)=180

Solution for (x-10)+(2x-123)+(1/2x+5)=180 equation:

(x-10)+(2x-123)+(1/2x+5)=180

We move all terms to the left:

(x-10)+(2x-123)+(1/2x+5)-(180)=0


Domain of the equation: 2x+5)!=0

x∈R


We get rid of parentheses

x+2x+1/2x-10-123+5-180=0

We multiply all the terms by the denominator


x*2x+2x*2x-10*2x-123*2x+5*2x-180*2x+1=0

Wy multiply elements

2x^2+4x^2-20x-246x+10x-360x+1=0

We add all the numbers together, and all the variables

6x^2-616x+1=0

a = 6; b = -616; c = +1;
Δ = b2-4ac
Δ = -6162-4·6·1
Δ = 379432
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:

$\sqrt{\Delta}=\sqrt{379432}=\sqrt{4*94858}=\sqrt{4}*\sqrt{94858}=2\sqrt{94858}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-616)-2\sqrt{94858}}{2*6}=\frac{616-2\sqrt{94858}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-616)+2\sqrt{94858}}{2*6}=\frac{616+2\sqrt{94858}}{12}
$