12m^2-20m+3=0

Solution for 12m^2-20m+3=0 equation:

12m^2-20m+3=0

a = 12; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·12·3
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*12}=\frac{4}{24}
=1/6
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*12}=\frac{36}{24}
=1+1/2
$