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m2-8m+10=0
We add all the numbers together, and all the variables
m^2-8m+10=0
a = 1; b = -8; c = +10;
Δ = b2-4ac
Δ = -82-4·1·10
Δ = 24
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{6}}{2*1}=\frac{8-2\sqrt{6}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{6}}{2*1}=\frac{8+2\sqrt{6}}{2} $
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