(10r+40)/(r^2+8r)=3

Solution for (10r+40)/(r^2+8r)=3 equation:

(10r+40)/(r^2+8r)=3

We move all terms to the left:

(10r+40)/(r^2+8r)-(3)=0


Domain of the equation: (r^2+8r)!=0

r∈R


We multiply all the terms by the denominator


(10r+40)-3*(r^2+8r)=0

We multiply parentheses

-3r^2+(10r+40)-24r=0

We get rid of parentheses

-3r^2+10r-24r+40=0

We add all the numbers together, and all the variables

-3r^2-14r+40=0

a = -3; b = -14; c = +40;
Δ = b2-4ac
Δ = -142-4·(-3)·40
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*-3}=\frac{-12}{-6}
=+2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*-3}=\frac{40}{-6}
=-6+2/3
$