Solve x^2+9x-3>0

$$x ^ { 2 } + 9 x - 3 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{93}-9}{2}\right)\cup \left(\frac{\sqrt{93}-9}{2},\infty\right)$$

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.

$$x^{2}+9x-3=0$$

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 1 for a, 9 for b, and -3 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{-9±\sqrt{93}}{2}$$

Solve the equation x=\frac{-9±\sqrt{93}}{2} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{93}-9}{2} x=\frac{-\sqrt{93}-9}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{\sqrt{93}-9}{2}\right)\left(x-\frac{-\sqrt{93}-9}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{93}-9}{2} and x-\frac{-\sqrt{93}-9}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{93}-9}{2} and x-\frac{-\sqrt{93}-9}{2} are both negative.

$$x-\frac{\sqrt{93}-9}{2}<0 x-\frac{-\sqrt{93}-9}{2}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{93}-9}{2}.

$$x<\frac{-\sqrt{93}-9}{2}$$

Consider the case when x-\frac{\sqrt{93}-9}{2} and x-\frac{-\sqrt{93}-9}{2} are both positive.

$$x-\frac{-\sqrt{93}-9}{2}>0 x-\frac{\sqrt{93}-9}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{93}-9}{2}.

$$x>\frac{\sqrt{93}-9}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{93}-9}{2}\text{; }x>\frac{\sqrt{93}-9}{2}$$