m+50=40(m2)

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Solution for m+50=40(m2) equation:



m+50=40(m2)
We move all terms to the left:
m+50-(40(m2))=0
determiningTheFunctionDomain m-40m2+50=0
We add all the numbers together, and all the variables
-40m^2+m+50=0
a = -40; b = 1; c = +50;
Δ = b2-4ac
Δ = 12-4·(-40)·50
Δ = 8001
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:
$\sqrt{\Delta}=\sqrt{8001}=\sqrt{9*889}=\sqrt{9}*\sqrt{889}=3\sqrt{889}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{889}}{2*-40}=\frac{-1-3\sqrt{889}}{-80} $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{889}}{2*-40}=\frac{-1+3\sqrt{889}}{-80} $

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