Solve 7x+5x^2+1>0

$$7 x + 5 x ^ { 2 } + 1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{29}-7}{10}\right)\cup \left(\frac{\sqrt{29}-7}{10},\infty\right)$$

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.

$$7x+5x^{2}+1=0$$

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 5 for a, 7 for b, and 1 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 5\times 1}}{2\times 5}$$

Do the calculations.

$$x=\frac{-7±\sqrt{29}}{10}$$

Solve the equation x=\frac{-7±\sqrt{29}}{10} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{29}-7}{10} x=\frac{-\sqrt{29}-7}{10}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-\frac{\sqrt{29}-7}{10}\right)\left(x-\frac{-\sqrt{29}-7}{10}\right)>0$$

For the product to be positive, x-\frac{\sqrt{29}-7}{10} and x-\frac{-\sqrt{29}-7}{10} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{29}-7}{10} and x-\frac{-\sqrt{29}-7}{10} are both negative.

$$x-\frac{\sqrt{29}-7}{10}<0 x-\frac{-\sqrt{29}-7}{10}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{29}-7}{10}.

$$x<\frac{-\sqrt{29}-7}{10}$$

Consider the case when x-\frac{\sqrt{29}-7}{10} and x-\frac{-\sqrt{29}-7}{10} are both positive.

$$x-\frac{-\sqrt{29}-7}{10}>0 x-\frac{\sqrt{29}-7}{10}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{29}-7}{10}.

$$x>\frac{\sqrt{29}-7}{10}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{29}-7}{10}\text{; }x>\frac{\sqrt{29}-7}{10}$$