Solve x^2-3x-3>0

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

Solve for x

$$x\in \left(-\infty,\frac{3-\sqrt{21}}{2}\right)\cup \left(\frac{\sqrt{21}+3}{2},\infty\right)$$

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.

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

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

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

Do the calculations.

$$x=\frac{3±\sqrt{21}}{2}$$

Solve the equation x=\frac{3±\sqrt{21}}{2} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{21}+3}{2} x=\frac{3-\sqrt{21}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{\sqrt{21}+3}{2}\right)\left(x-\frac{3-\sqrt{21}}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{21}+3}{2} and x-\frac{3-\sqrt{21}}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{21}+3}{2} and x-\frac{3-\sqrt{21}}{2} are both negative.

$$x-\frac{\sqrt{21}+3}{2}<0 x-\frac{3-\sqrt{21}}{2}<0$$

The solution satisfying both inequalities is x<\frac{3-\sqrt{21}}{2}.

$$x<\frac{3-\sqrt{21}}{2}$$

Consider the case when x-\frac{\sqrt{21}+3}{2} and x-\frac{3-\sqrt{21}}{2} are both positive.

$$x-\frac{3-\sqrt{21}}{2}>0 x-\frac{\sqrt{21}+3}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{21}+3}{2}.

$$x>\frac{\sqrt{21}+3}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{3-\sqrt{21}}{2}\text{; }x>\frac{\sqrt{21}+3}{2}$$