(2x(x-8))/(x-4)^2=0

Solution for (2x(x-8))/(x-4)^2=0 equation:

(2x(x-8))/(x-4)^2=0


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

x∈R


We multiply all the terms by the denominator


(2x(x-8))=0


We calculate terms in parentheses: +(2x(x-8)), so:

2x(x-8)

We multiply parentheses

2x^2-16x

Back to the equation:

+(2x^2-16x)


We get rid of parentheses

2x^2-16x=0

a = 2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·2·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*2}=\frac{32}{4}
=8
$