4x+32=8(1/2x-4)

Solution for 4x+32=8(1/2x-4) equation:

4x+32=8(1/2x-4)

We move all terms to the left:

4x+32-(8(1/2x-4))=0


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

x∈R


We multiply all the terms by the denominator


4x*2x+32*2x-4))-(8(1-4))=0

We add all the numbers together, and all the variables

4x*2x+32*2x-4))-(8(-3))=0

We add all the numbers together, and all the variables

4x*2x+32*2x=0

Wy multiply elements

8x^2+64x=0

a = 8; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·8·0
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*8}=\frac{-128}{16}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*8}=\frac{0}{16}
=0
$