3r^2+r+2=2

Solution for 3r^2+r+2=2 equation:

3r^2+r+2=2

We move all terms to the left:

3r^2+r+2-(2)=0

We add all the numbers together, and all the variables

3r^2+r=0

a = 3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·3·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*3}=\frac{-2}{6}
=-1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*3}=\frac{0}{6}
=0
$