4m^2-18m+18=0

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

4m^2-18m+18=0

a = 4; b = -18; c = +18;
Δ = b2-4ac
Δ = -182-4·4·18
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*4}=\frac{12}{8}
=1+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*4}=\frac{24}{8}
=3
$