7m^2-18m+7=0

Solution for 7m^2-18m+7=0 equation:

7m^2-18m+7=0

a = 7; b = -18; c = +7;
Δ = b2-4ac
Δ = -182-4·7·7
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8\sqrt{2}}{2*7}=\frac{18-8\sqrt{2}}{14}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8\sqrt{2}}{2*7}=\frac{18+8\sqrt{2}}{14}
$