(4/x-4)+(5x-20)=-5

Solution for (4/x-4)+(5x-20)=-5 equation:

(4/x-4)+(5x-20)=-5

We move all terms to the left:

(4/x-4)+(5x-20)-(-5)=0


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

x∈R


We add all the numbers together, and all the variables

(4/x-4)+(5x-20)+5=0

We get rid of parentheses

4/x+5x-4-20+5=0

We multiply all the terms by the denominator


5x*x-4*x-20*x+5*x+4=0

We add all the numbers together, and all the variables

-19x+5x*x+4=0

Wy multiply elements

5x^2-19x+4=0

a = 5; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·5·4
Δ = 281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{281}}{2*5}=\frac{19-\sqrt{281}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{281}}{2*5}=\frac{19+\sqrt{281}}{10}
$