Solve -3(x-9)(2+x)>0

$$-3(x-9)(2+x) > 0$$

Solve for x

$$x\in \left(-2,9\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -3 by x-9.

$$\left(-3x+27\right)\left(2+x\right)>0$$

Use the distributive property to multiply -3x+27 by 2+x and combine like terms.

$$21x-3x^{2}+54>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 21x-3x^{2}+54 positive. Since -1 is negative, the inequality direction is changed.

$$-21x+3x^{2}-54<0$$

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.

$$-21x+3x^{2}-54=0$$

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 3 for a, -21 for b, and -54 for c in the quadratic formula.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\left(-54\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{21±33}{6}$$

Solve the equation x=\frac{21±33}{6} when ± is plus and when ± is minus.

$$x=9 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-9\right)\left(x+2\right)<0$$

For the product to be negative, x-9 and x+2 have to be of the opposite signs. Consider the case when x-9 is positive and x+2 is negative.

$$x-9>0 x+2<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+2 is positive and x-9 is negative.

$$x+2>0 x-9<0$$

The solution satisfying both inequalities is x\in \left(-2,9\right).

$$x\in \left(-2,9\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-2,9\right)$$