3(2r-1)-r+8=5(r+1)r=

Solution for 3(2r-1)-r+8=5(r+1)r= equation:

3(2r-1)-r+8=5(r+1)r=

We move all terms to the left:

3(2r-1)-r+8-(5(r+1)r)=0

We add all the numbers together, and all the variables

-1r+3(2r-1)-(5(r+1)r)+8=0

We multiply parentheses

-1r+6r-(5(r+1)r)-3+8=0


We calculate terms in parentheses: -(5(r+1)r), so:

5(r+1)r

We multiply parentheses

5r^2+5r

Back to the equation:

-(5r^2+5r)


We add all the numbers together, and all the variables

5r-(5r^2+5r)+5=0

We get rid of parentheses

-5r^2+5r-5r+5=0

We add all the numbers together, and all the variables

-5r^2+5=0

a = -5; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-5)·5
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-5}=\frac{-10}{-10}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-5}=\frac{10}{-10}
=-1
$