(1/2)(2x+2+2x+4)=28

Solution for (1/2)(2x+2+2x+4)=28 equation:

(1/2)(2x+2+2x+4)=28

We move all terms to the left:

(1/2)(2x+2+2x+4)-(28)=0


Domain of the equation: 2)(2x+2+2x+4)!=0

We move all terms containing x to the left, all other terms to the right

2)(2x+2x+4)!=-2

x∈R


We add all the numbers together, and all the variables

(+1/2)(4x+6)-28=0

We multiply parentheses ..

(+4x^2+1/2*6)-28=0

We multiply all the terms by the denominator


(+4x^2+1-28*2*6)=0

We get rid of parentheses

4x^2+1-28*2*6=0

We add all the numbers together, and all the variables

4x^2-335=0

a = 4; b = 0; c = -335;
Δ = b2-4ac
Δ = 02-4·4·(-335)
Δ = 5360
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:

$\sqrt{\Delta}=\sqrt{5360}=\sqrt{16*335}=\sqrt{16}*\sqrt{335}=4\sqrt{335}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{335}}{2*4}=\frac{0-4\sqrt{335}}{8}
=-\frac{4\sqrt{335}}{8}
=-\frac{\sqrt{335}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{335}}{2*4}=\frac{0+4\sqrt{335}}{8}
=\frac{4\sqrt{335}}{8}
=\frac{\sqrt{335}}{2}
$