Solve -(x+2)(x-5)<0

$$- ( x + 2 ) ( x - 5 ) < 0$$

Solve for x

$$x\in \left(-\infty,-2\right)\cup \left(5,\infty\right)$$

Steps Using the Quadratic Formula

To find the opposite of x+2, find the opposite of each term.

$$\left(-x-2\right)\left(x-5\right)<0$$

Apply the distributive property by multiplying each term of -x-2 by each term of x-5.

$$-x^{2}+5x-2x+10<0$$

Combine 5x and -2x to get 3x.

$$-x^{2}+3x+10<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+3x+10 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-3x-10>0$$

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.

$$x^{2}-3x-10=0$$

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, -3 for b, and -10 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{3±7}{2}$$

Solve the equation x=\frac{3±7}{2} when ± is plus and when ± is minus.

$$x=5 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x+2\right)>0$$

For the product to be positive, x-5 and x+2 have to be both negative or both positive. Consider the case when x-5 and x+2 are both negative.

$$x-5<0 x+2<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when x-5 and x+2 are both positive.

$$x+2>0 x-5>0$$

The solution satisfying both inequalities is x>5.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x>5$$