Solve 2(x-2)(x+3)<0

$$2 ( x - 2 ) ( x + 3 ) < 0$$

Solve for x

$$x\in \left(-3,2\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x-2.

$$\left(2x-4\right)\left(x+3\right)<0$$

Use the distributive property to multiply 2x-4 by x+3 and combine like terms.

$$2x^{2}+2x-12<0$$

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.

$$2x^{2}+2x-12=0$$

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 2 for a, 2 for b, and -12 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±10}{4}$$

Solve the equation x=\frac{-2±10}{4} when ± is plus and when ± is minus.

$$x=2 x=-3$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be negative, x-2 and x+3 have to be of the opposite signs. Consider the case when x-2 is positive and x+3 is negative.

$$x-2>0 x+3<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+3 is positive and x-2 is negative.

$$x+3>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(-3,2\right).

$$x\in \left(-3,2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-3,2\right)$$