1/2x+3(2x+2)=20-x

Solution for 1/2x+3(2x+2)=20-x equation:

1/2x+3(2x+2)=20-x

We move all terms to the left:

1/2x+3(2x+2)-(20-x)=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+3(2x+2)-(-1x+20)=0

We multiply parentheses

1/2x+6x-(-1x+20)+6=0

We get rid of parentheses

1/2x+6x+1x-20+6=0

We multiply all the terms by the denominator


6x*2x+1x*2x-20*2x+6*2x+1=0

Wy multiply elements

12x^2+2x^2-40x+12x+1=0

We add all the numbers together, and all the variables

14x^2-28x+1=0

a = 14; b = -28; c = +1;
Δ = b2-4ac
Δ = -282-4·14·1
Δ = 728
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:

$\sqrt{\Delta}=\sqrt{728}=\sqrt{4*182}=\sqrt{4}*\sqrt{182}=2\sqrt{182}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{182}}{2*14}=\frac{28-2\sqrt{182}}{28}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{182}}{2*14}=\frac{28+2\sqrt{182}}{28}
$