\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{16x}\text{ }=\text{ }\mathbf{0}~~~~~~\]

Solution:-

\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{16x}\text{ }=\text{ }\mathbf{0}~~~~~~\]

Take out common in each terms,

$4x(x + 4) = 0$

Equate both to zero,

$4x = 0$,

$x + 4 = 0$

$x = 0$

$x = – 4$