\[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{x}\text{ }-\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]

Solution:-

\[2{{x}^{2}}~-x\text{ }-\text{ }6\text{ }=\text{ }0~~~~~~\]

Divided by \[2\]for both side of each term we get,

\[2{{x}^{2}}/2\text{ }-x/2\text{ }-\text{ }6/2\text{ }=\text{ }0\]

\[{{x}^{2}}~-2x\text{ }+\text{ }\left( 3/2 \right)x\text{ }-\text{ }3\text{ }=\text{ }0\]

\[2{{x}^{2}}/2\text{ }-\text{ }x/2\text{ }-\text{ }6/2\text{ }=\text{ }0\]

Take out common in each terms,

$x(x – 2) +3/2 (x – 2) = 0$

$(x – 2) + (x + (3/2)) = 0$

Equate both to zero,

$x – 2 = 0$

 $x + 3/2 = 0$

$x = 2$

 $x = – 3/2$