\[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{9}\text{ }=\text{ }\mathbf{0}\]

Solution:-

\[2{{x}^{2}}-3x-9\text{ }=\text{ }0\]

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

\[(2{{x}^{2}}/2)\text{ }-\left( 3x/2 \right)\text{ }-\left( 9/2 \right)\text{ }=\text{ }0/2\]

\[{{x}^{2}}~-\text{ }3x/2\text{ }-9/2\text{ }=\text{ }0\]

\[{{x}^{2}}~-\text{ }3x\text{ }+\text{ }\left( 3/2 \right)x\text{ }-\text{ }9/2\text{ }=\text{ }0\]

Take out common in each terms,

$x(x – 3) + (3/2) (x – 3) = 0$

$(x + 3/2) (x – 3) = 0$

Equate both to zero,

$x + 3/2 = 0$

 $x – 3 = 0$

$x = -3/2$

$x = 3$