(8x+32)=(2x+12)+1/2x

Solution for (8x+32)=(2x+12)+1/2x equation:

(8x+32)=(2x+12)+1/2x

We move all terms to the left:

(8x+32)-((2x+12)+1/2x)=0


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

8x-((2x+12)+1/2x)+32=0

We multiply all the terms by the denominator


8x*2x)-((2x+12)+32*2x)+1=0

Wy multiply elements

16x^2+64x=0

a = 16; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·16·0
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*16}=\frac{-128}{32}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*16}=\frac{0}{32}
=0
$