r^-5r=r(r+1)

Solution for r^-5r=r(r+1) equation:

r^-5r=r(r+1)

We move all terms to the left:

r^-5r-(r(r+1))=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(r(r+1)), so:

r(r+1)

We multiply parentheses

r^2+r

Back to the equation:

-(r^2+r)


We get rid of parentheses

-r^2-4r-r=0

We add all the numbers together, and all the variables

-1r^2-5r=0

a = -1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-1}=\frac{0}{-2}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-1}=\frac{10}{-2}
=-5
$