(1/2x+2)+1/2x+x+4x=122

Solution for (1/2x+2)+1/2x+x+4x=122 equation:

(1/2x+2)+1/2x+x+4x=122

We move all terms to the left:

(1/2x+2)+1/2x+x+4x-(122)=0


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

x∈R



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

5x+(1/2x+2)+1/2x-122=0

We get rid of parentheses

5x+1/2x+1/2x+2-122=0

We multiply all the terms by the denominator


5x*2x+2*2x-122*2x+1+1=0

We add all the numbers together, and all the variables

5x*2x+2*2x-122*2x+2=0

Wy multiply elements

10x^2+4x-244x+2=0

We add all the numbers together, and all the variables

10x^2-240x+2=0

a = 10; b = -240; c = +2;
Δ = b2-4ac
Δ = -2402-4·10·2
Δ = 57520
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:

$\sqrt{\Delta}=\sqrt{57520}=\sqrt{16*3595}=\sqrt{16}*\sqrt{3595}=4\sqrt{3595}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-4\sqrt{3595}}{2*10}=\frac{240-4\sqrt{3595}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+4\sqrt{3595}}{2*10}=\frac{240+4\sqrt{3595}}{20}
$