\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{13x}\text{ }-\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\]

Solution:-

\[4{{x}^{2}}~-13x\text{ }-12\text{ }=\text{ }0\]

Divided by \[4\]for both side of each term we get,

\[4{{x}^{2}}/4\text{ }-13x/4\text{ }-\text{ }12/4\text{ }=\text{ }0/4\]

\[{{x}^{2}}~-13x/4\text{ }-\text{ }3\text{ }=\text{ }0\]

\[{{x}^{2}}~-\text{ }4x\text{ }+\text{ }3x/4\text{ }-\text{ }3\text{ }=\text{ }0\]

Take out common in each terms,

$x(x – 4) + 3/4(x – 4) = 0$

$(x – 4) (x + 3/4) = 0$

Equate both to zero,

$(x – 4) = 0$

$x + 3/4 = 0$

$4X = 4$

$x = -3/4$