m2+18m+45=0

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Solution for m2+18m+45=0 equation:



m2+18m+45=0
We add all the numbers together, and all the variables
m^2+18m+45=0
a = 1; b = 18; c = +45;
Δ = b2-4ac
Δ = 182-4·1·45
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12}{2*1}=\frac{-30}{2} =-15 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*1}=\frac{-6}{2} =-3 $

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