0.3r-0.06r^2=0

Solution for 0.3r-0.06r^2=0 equation:

0.3r-0.06r^2=0

a = -0.06; b = 0.3; c = 0;
Δ = b2-4ac
Δ = 0.32-4·(-0.06)·0
Δ = 0.09
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{-(0.3)-\sqrt{0.09}}{2*-0.06}=\frac{-0.3-\sqrt{0.09}}{-0.12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.09}}{2*-0.06}=\frac{-0.3+\sqrt{0.09}}{-0.12}
$