(1/2x+20)+(1/2x+20)=3x-12

Solution for (1/2x+20)+(1/2x+20)=3x-12 equation:

(1/2x+20)+(1/2x+20)=3x-12

We move all terms to the left:

(1/2x+20)+(1/2x+20)-(3x-12)=0


Domain of the equation: 2x+20)!=0

x∈R


We get rid of parentheses

1/2x+1/2x-3x+20+20+12=0

We multiply all the terms by the denominator


-3x*2x+20*2x+20*2x+12*2x+1+1=0

We add all the numbers together, and all the variables

-3x*2x+20*2x+20*2x+12*2x+2=0

Wy multiply elements

-6x^2+40x+40x+24x+2=0

We add all the numbers together, and all the variables

-6x^2+104x+2=0

a = -6; b = 104; c = +2;
Δ = b2-4ac
Δ = 1042-4·(-6)·2
Δ = 10864
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:

$\sqrt{\Delta}=\sqrt{10864}=\sqrt{16*679}=\sqrt{16}*\sqrt{679}=4\sqrt{679}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-4\sqrt{679}}{2*-6}=\frac{-104-4\sqrt{679}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+4\sqrt{679}}{2*-6}=\frac{-104+4\sqrt{679}}{-12}
$