Solve 5(x-13)(x+24)<0

$$5 ( x - 13 ) ( x + 24 ) < 0$$

Solve for x

$$x\in \left(-24,13\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5 by x-13.

$$\left(5x-65\right)\left(x+24\right)<0$$

Use the distributive property to multiply 5x-65 by x+24 and combine like terms.

$$5x^{2}+55x-1560<0$$

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.

$$5x^{2}+55x-1560=0$$

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 5 for a, 55 for b, and -1560 for c in the quadratic formula.

$$x=\frac{-55±\sqrt{55^{2}-4\times 5\left(-1560\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{-55±185}{10}$$

Solve the equation x=\frac{-55±185}{10} when ± is plus and when ± is minus.

$$x=13 x=-24$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-13\right)\left(x+24\right)<0$$

For the product to be negative, x-13 and x+24 have to be of the opposite signs. Consider the case when x-13 is positive and x+24 is negative.

$$x-13>0 x+24<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+24 is positive and x-13 is negative.

$$x+24>0 x-13<0$$

The solution satisfying both inequalities is x\in \left(-24,13\right).

$$x\in \left(-24,13\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-24,13\right)$$