103=r+.05r^2

Solution for 103=r+.05r^2 equation:

103=r+.05r^2

We move all terms to the left:

103-(r+.05r^2)=0

We get rid of parentheses

-.05r^2-r+103=0

We add all the numbers together, and all the variables

-0.05r^2-1r+103=0

a = -0.05; b = -1; c = +103;
Δ = b2-4ac
Δ = -12-4·(-0.05)·103
Δ = 21.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{21.6}}{2*-0.05}=\frac{1-\sqrt{21.6}}{-0.1}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{21.6}}{2*-0.05}=\frac{1+\sqrt{21.6}}{-0.1}
$