Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-9$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=2\left(-9\right)=-18$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-18$.
$$-1,18$$ $$-2,9$$ $$-3,6$$
Calculate the sum for each pair.
$$-1+18=17$$ $$-2+9=7$$ $$-3+6=3$$
The solution is the pair that gives sum $3$.
$$a=-3$$ $$b=6$$
Rewrite $2x^{2}+3x-9$ as $\left(2x^{2}-3x\right)+\left(6x-9\right)$.
$$\left(2x^{2}-3x\right)+\left(6x-9\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(2x-3\right)+3\left(2x-3\right)$$
Factor out common term $2x-3$ by using distributive property.
