m^2-9m+15=0

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Solution for m^2-9m+15=0 equation: 



m^2-9m+15=0

a = 1; b = -9; c = +15;
Δ = b2-4ac
Δ = -92-4·1·15
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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

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