25m+20m^2=0

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Solution for 25m+20m^2=0 equation: 



25m+20m^2=0

a = 20; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·20·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*20}=\frac{-50}{40}
=-1+1/4
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*20}=\frac{0}{40}
=0
$

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