Solve 3x^2+5x+1>0

$$3 x ^ { 2 } + 5 x + 1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{13}-5}{6}\right)\cup \left(\frac{\sqrt{13}-5}{6},\infty\right)$$

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.

$$3x^{2}+5x+1=0$$

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, 5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$x=\frac{-5±\sqrt{13}}{6}$$

Solve the equation x=\frac{-5±\sqrt{13}}{6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{13}-5}{6} x=\frac{-\sqrt{13}-5}{6}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\frac{\sqrt{13}-5}{6}\right)\left(x-\frac{-\sqrt{13}-5}{6}\right)>0$$

For the product to be positive, x-\frac{\sqrt{13}-5}{6} and x-\frac{-\sqrt{13}-5}{6} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{13}-5}{6} and x-\frac{-\sqrt{13}-5}{6} are both negative.

$$x-\frac{\sqrt{13}-5}{6}<0 x-\frac{-\sqrt{13}-5}{6}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{13}-5}{6}.

$$x<\frac{-\sqrt{13}-5}{6}$$

Consider the case when x-\frac{\sqrt{13}-5}{6} and x-\frac{-\sqrt{13}-5}{6} are both positive.

$$x-\frac{-\sqrt{13}-5}{6}>0 x-\frac{\sqrt{13}-5}{6}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{13}-5}{6}.

$$x>\frac{\sqrt{13}-5}{6}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{13}-5}{6}\text{; }x>\frac{\sqrt{13}-5}{6}$$