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m2-6m-1=0
We add all the numbers together, and all the variables
m^2-6m-1=0
a = 1; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·1·(-1)
Δ = 40
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{10}}{2*1}=\frac{6-2\sqrt{10}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{10}}{2*1}=\frac{6+2\sqrt{10}}{2} $
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