Solve 4x^2-13x+3>0

$$4 x ^ { 2 } - 13 x + 3 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1}{4}\right)\cup \left(3,\infty\right)$$

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.

$$4x^{2}-13x+3=0$$

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 4 for a, -13 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Do the calculations.

$$x=\frac{13±11}{8}$$

Solve the equation x=\frac{13±11}{8} when ± is plus and when ± is minus.

$$x=3 x=\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-3\right)\left(x-\frac{1}{4}\right)>0$$

For the product to be positive, x-3 and x-\frac{1}{4} have to be both negative or both positive. Consider the case when x-3 and x-\frac{1}{4} are both negative.

$$x-3<0 x-\frac{1}{4}<0$$

The solution satisfying both inequalities is x<\frac{1}{4}.

$$x<\frac{1}{4}$$

Consider the case when x-3 and x-\frac{1}{4} are both positive.

$$x-\frac{1}{4}>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{4}\text{; }x>3$$