12=0.0023r^2+r

Solution for 12=0.0023r^2+r equation:

12=0.0023r^2+r

We move all terms to the left:

12-(0.0023r^2+r)=0

We get rid of parentheses

-0.0023r^2-r+12=0

We add all the numbers together, and all the variables

-0.0023r^2-1r+12=0

a = -0.0023; b = -1; c = +12;
Δ = b2-4ac
Δ = -12-4·(-0.0023)·12
Δ = 1.1104
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{1.1104}}{2*-0.0023}=\frac{1-\sqrt{1.1104}}{-0.0046}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1.1104}}{2*-0.0023}=\frac{1+\sqrt{1.1104}}{-0.0046}
$