Solve 2x^2-25x+12>0

$$2 x ^ { 2 } - 25 x + 12 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1}{2}\right)\cup \left(12,\infty\right)$$

Steps Using the Quadratic Formula

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.

$$2x^{2}-25x+12=0$$

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, -25 for b, and 12 for c in the quadratic formula.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 2\times 12}}{2\times 2}$$

Do the calculations.

$$x=\frac{25±23}{4}$$

Solve the equation x=\frac{25±23}{4} when ± is plus and when ± is minus.

$$x=12 x=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-12\right)\left(x-\frac{1}{2}\right)>0$$

For the product to be positive, x-12 and x-\frac{1}{2} have to be both negative or both positive. Consider the case when x-12 and x-\frac{1}{2} are both negative.

$$x-12<0 x-\frac{1}{2}<0$$

The solution satisfying both inequalities is x<\frac{1}{2}.

$$x<\frac{1}{2}$$

Consider the case when x-12 and x-\frac{1}{2} are both positive.

$$x-\frac{1}{2}>0 x-12>0$$

The solution satisfying both inequalities is x>12.

$$x>12$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{2}\text{; }x>12$$