-100+33/(1+r)(r)=0

Solution for -100+33/(1+r)(r)=0 equation:

-100+33/(1+r)(r)=0


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

r∈R


We add all the numbers together, and all the variables

33/(r+1)r-100=0

We multiply all the terms by the denominator


-100*(r+1)r+33=0

We multiply parentheses

-100r^2-100r+33=0

a = -100; b = -100; c = +33;
Δ = b2-4ac
Δ = -1002-4·(-100)·33
Δ = 23200
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:

$\sqrt{\Delta}=\sqrt{23200}=\sqrt{400*58}=\sqrt{400}*\sqrt{58}=20\sqrt{58}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-20\sqrt{58}}{2*-100}=\frac{100-20\sqrt{58}}{-200}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+20\sqrt{58}}{2*-100}=\frac{100+20\sqrt{58}}{-200}
$