18m^2+12m-6=0

Solution for 18m^2+12m-6=0 equation:

18m^2+12m-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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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