3m^2-18m+4=0

Solution for 3m^2-18m+4=0 equation:

3m^2-18m+4=0

a = 3; b = -18; c = +4;
Δ = b2-4ac
Δ = -182-4·3·4
Δ = 276
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{69}}{2*3}=\frac{18-2\sqrt{69}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{69}}{2*3}=\frac{18+2\sqrt{69}}{6}
$