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1/(x-3)-4/x-3-5=0

Domain of the equation: (x-3)!=0

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

x!=3

x∈R

Domain of the equation: x!=0We add all the numbers together, and all the variables

x∈R

1/(x-3)-4/x-8=0

We calculate fractions

x/(x^2-3x)+(-4x+12)/(x^2-3x)-8=0

We multiply all the terms by the denominator

x+(-4x+12)-8*(x^2-3x)=0

We multiply parentheses

-8x^2+x+(-4x+12)+24x=0

We get rid of parentheses

-8x^2+x-4x+24x+12=0

We add all the numbers together, and all the variables

-8x^2+21x+12=0

a = -8; b = 21; c = +12;

Δ = b^{2}-4ac

Δ = 21^{2}-4·(-8)·12

Δ = 825

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}$

The end solution:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-5\sqrt{33}}{2*-8}=\frac{-21-5\sqrt{33}}{-16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+5\sqrt{33}}{2*-8}=\frac{-21+5\sqrt{33}}{-16} $

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