0.02r+0.01=0.15r^2

Solution for 0.02r+0.01=0.15r^2 equation:

0.02r+0.01=0.15r^2

We move all terms to the left:

0.02r+0.01-(0.15r^2)=0

We get rid of parentheses

-0.15r^2+0.02r+0.01=0

a = -0.15; b = 0.02; c = +0.01;
Δ = b2-4ac
Δ = 0.022-4·(-0.15)·0.01
Δ = 0.0064
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{-(0.02)-\sqrt{0.0064}}{2*-0.15}=\frac{-0.02-\sqrt{0.0064}}{-0.3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.02)+\sqrt{0.0064}}{2*-0.15}=\frac{-0.02+\sqrt{0.0064}}{-0.3}
$