(2x^2-8x)/(x3-8)=0

Solution for (2x^2-8x)/(x3-8)=0 equation:

(2x^2-8x)/(x3-8)=0


Domain of the equation: (x3-8)!=0

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

x3!=8

x^3!=8/

x^3!=3√1/0

x!=1

x∈R


We add all the numbers together, and all the variables

(2x^2-8x)/(+x^3-8)=0

We multiply all the terms by the denominator


(2x^2-8x)=0

We get rid of parentheses

2x^2-8x=0

a = 2; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*2}=\frac{16}{4}
=4
$