m^2+48m=0

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Solution for m^2+48m=0 equation: 



m^2+48m=0

a = 1; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·1·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*1}=\frac{-96}{2}
=-48
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*1}=\frac{0}{2}
=0
$

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