m^2+4m+m^2+4m=0

Solution for m^2+4m+m^2+4m=0 equation:

m^2+4m+m^2+4m=0

We add all the numbers together, and all the variables

2m^2+8m=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$