r-4/5r=1/5r+1

Solution for r-4/5r=1/5r+1 equation:

r-4/5r=1/5r+1

We move all terms to the left:

r-4/5r-(1/5r+1)=0


Domain of the equation: 5r!=0

r!=0/5

r!=0

r∈R



Domain of the equation: 5r+1)!=0

r∈R


We get rid of parentheses

r-4/5r-1/5r-1=0

We multiply all the terms by the denominator


r*5r-1*5r-4-1=0

We add all the numbers together, and all the variables

r*5r-1*5r-5=0

Wy multiply elements

5r^2-5r-5=0

a = 5; b = -5; c = -5;
Δ = b2-4ac
Δ = -52-4·5·(-5)
Δ = 125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{5}}{2*5}=\frac{5-5\sqrt{5}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{5}}{2*5}=\frac{5+5\sqrt{5}}{10}
$