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x^{2}-9x+18=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 1\times 18}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -9 for b, and 18 for c in the quadratic formula.
x=\frac{9±3}{2}
Do the calculations.
x=6 x=3
Solve the equation x=\frac{9±3}{2} when ± is plus and when ± is minus.
\left(x-6\right)\left(x-3\right)<0
Rewrite the inequality by using the obtained solutions.
x-6>0 x-3<0
For the product to be negative, x-6 and x-3 have to be of the opposite signs. Consider the case when x-6 is positive and x-3 is negative.
x\in \emptyset
This is false for any x.
x-3>0 x-6<0
Consider the case when x-3 is positive and x-6 is negative.
x\in \left(3,6\right)
The solution satisfying both inequalities is x\in \left(3,6\right).
x\in \left(3,6\right)
The final solution is the union of the obtained solutions.