18r2+12r=6

Solution for 18r2+12r=6 equation:

18r^2+12r=6

We move all terms to the left:

18r^2+12r-(6)=0

a = 18; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·18·(-6)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*18}=\frac{-36}{36}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*18}=\frac{12}{36}
=1/3
$