Solve x^2-14x+45<0

$$x ^ { 2 } - 14 x + 45 < 0$$

Solve for x

$$x\in \left(5,9\right)$$

Steps Using the Quadratic Formula

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.

$$x^{2}-14x+45=0$$

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, -14 for b, and 45 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\times 45}}{2}$$

Do the calculations.

$$x=\frac{14±4}{2}$$

Solve the equation x=\frac{14±4}{2} when ± is plus and when ± is minus.

$$x=9 x=5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-9\right)\left(x-5\right)<0$$

For the product to be negative, x-9 and x-5 have to be of the opposite signs. Consider the case when x-9 is positive and x-5 is negative.

$$x-9>0 x-5<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-5 is positive and x-9 is negative.

$$x-5>0 x-9<0$$

The solution satisfying both inequalities is x\in \left(5,9\right).

$$x\in \left(5,9\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(5,9\right)$$