m^2-21m+108=0

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Solution for m^2-21m+108=0 equation: 



m^2-21m+108=0

a = 1; b = -21; c = +108;
Δ = b2-4ac
Δ = -212-4·1·108
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3}{2*1}=\frac{18}{2}
=9
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3}{2*1}=\frac{24}{2}
=12
$

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