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m2-16m+20=0
We add all the numbers together, and all the variables
m^2-16m+20=0
a = 1; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·1·20
Δ = 176
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{11}}{2*1}=\frac{16-4\sqrt{11}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{11}}{2*1}=\frac{16+4\sqrt{11}}{2} $
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