(x+x+32)/2+(2/3)x=56

Solution for (x+x+32)/2+(2/3)x=56 equation:

(x+x+32)/2+(2/3)x=56

We move all terms to the left:

(x+x+32)/2+(2/3)x-(56)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(2x+32)/2+(+2/3)x-56=0

We multiply parentheses

2x^2+(2x+32)/2-56=0

We multiply all the terms by the denominator


2x^2*2+(2x+32)-56*2=0

We add all the numbers together, and all the variables

2x^2*2+(2x+32)-112=0

Wy multiply elements

4x^2+(2x+32)-112=0

We get rid of parentheses

4x^2+2x+32-112=0

We add all the numbers together, and all the variables

4x^2+2x-80=0

a = 4; b = 2; c = -80;
Δ = b2-4ac
Δ = 22-4·4·(-80)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{321}}{2*4}=\frac{-2-2\sqrt{321}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{321}}{2*4}=\frac{-2+2\sqrt{321}}{8}
$