Solve 2(x-2)(x-7)>0

$$2 ( x - 2 ) ( x - 7 ) > 0$$

Solve for x

$$x\in \left(-\infty,2\right)\cup \left(7,\infty\right)$$

Steps Using the Quadratic Formula

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

$$\left(2x-4\right)\left(x-7\right)>0$$

Use the distributive property to multiply 2x-4 by x-7 and combine like terms.

$$2x^{2}-18x+28>0$$

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.

$$2x^{2}-18x+28=0$$

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, -18 for b, and 28 for c in the quadratic formula.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 28}}{2\times 2}$$

Do the calculations.

$$x=\frac{18±10}{4}$$

Solve the equation x=\frac{18±10}{4} when ± is plus and when ± is minus.

$$x=7 x=2$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-7\right)\left(x-2\right)>0$$

For the product to be positive, x-7 and x-2 have to be both negative or both positive. Consider the case when x-7 and x-2 are both negative.

$$x-7<0 x-2<0$$

The solution satisfying both inequalities is x<2.

$$x<2$$

Consider the case when x-7 and x-2 are both positive.

$$x-2>0 x-7>0$$

The solution satisfying both inequalities is x>7.

$$x>7$$

The final solution is the union of the obtained solutions.

$$x<2\text{; }x>7$$