20m^2+18m-18=0

Solution for 20m^2+18m-18=0 equation:

20m^2+18m-18=0

a = 20; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·20·(-18)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*20}=\frac{-60}{40}
=-1+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*20}=\frac{24}{40}
=3/5
$