1/2x+(x-10)+(x-20)+40=360

Solution for 1/2x+(x-10)+(x-20)+40=360 equation:

1/2x+(x-10)+(x-20)+40=360

We move all terms to the left:

1/2x+(x-10)+(x-20)+40-(360)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

1/2x+(x-10)+(x-20)-320=0

We get rid of parentheses

1/2x+x+x-10-20-320=0

We multiply all the terms by the denominator


x*2x+x*2x-10*2x-20*2x-320*2x+1=0

Wy multiply elements

2x^2+2x^2-20x-40x-640x+1=0

We add all the numbers together, and all the variables

4x^2-700x+1=0

a = 4; b = -700; c = +1;
Δ = b2-4ac
Δ = -7002-4·4·1
Δ = 489984
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:

$\sqrt{\Delta}=\sqrt{489984}=\sqrt{256*1914}=\sqrt{256}*\sqrt{1914}=16\sqrt{1914}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-700)-16\sqrt{1914}}{2*4}=\frac{700-16\sqrt{1914}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-700)+16\sqrt{1914}}{2*4}=\frac{700+16\sqrt{1914}}{8}
$