Verified Answer :

iloveonedirection

\(4x^2+16x=0\)

\(\frac{4x^2+16x}{4} =\frac{0}{4}\)

\(x^2+4x=0\)

\(x(x+4)=0\)

\(x=0\)

\(x+4=0\)

\(x+4-4=0-4\)

\(x=-4\)


\(x=-4,0\)

AL2006

Divide each side of the equation by 4 :

x² + 4x = 0

Factor the left side:

x (x + 4) = 0

This equation is true if x=0 or if x=-4 , so those are the two solutions.

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