(x(4-3x)+4(x^2+1))/x^2-4=0

Solution for (x(4-3x)+4(x^2+1))/x^2-4=0 equation:

(x(4-3x)+4(x^2+1))/x^2-4=0


Domain of the equation: x^2!=0

x^2!=0/

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

(x(-3x+4)+4(x^2+1))/x^2-4=0

We multiply all the terms by the denominator


(x(-3x+4)+4(x^2+1))-4*x^2=0


We calculate terms in parentheses: +(x(-3x+4)+4(x^2+1)), so:

x(-3x+4)+4(x^2+1)

We multiply parentheses

-3x^2+4x^2+4x+4

We add all the numbers together, and all the variables

x^2+4x+4

Back to the equation:

+(x^2+4x+4)


We add all the numbers together, and all the variables

-4x^2+(x^2+4x+4)=0

We get rid of parentheses

-4x^2+x^2+4x+4=0

We add all the numbers together, and all the variables

-3x^2+4x+4=0

a = -3; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·(-3)·4
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{-(4)-8}{2*-3}=\frac{-12}{-6}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-3}=\frac{4}{-6}
=-2/3
$