Solve 3x^2-7x+2>0

$$3 x ^ { 2 } - 7 x + 2 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1}{3}\right)\cup \left(2,\infty\right)$$

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.

$$3x^{2}-7x+2=0$$

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, -7 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Do the calculations.

$$x=\frac{7±5}{6}$$

Solve the equation x=\frac{7±5}{6} when ± is plus and when ± is minus.

$$x=2 x=\frac{1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-2\right)\left(x-\frac{1}{3}\right)>0$$

For the product to be positive, x-2 and x-\frac{1}{3} have to be both negative or both positive. Consider the case when x-2 and x-\frac{1}{3} are both negative.

$$x-2<0 x-\frac{1}{3}<0$$

The solution satisfying both inequalities is x<\frac{1}{3}.

$$x<\frac{1}{3}$$

Consider the case when x-2 and x-\frac{1}{3} are both positive.

$$x-\frac{1}{3}>0 x-2>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{3}\text{; }x>2$$