Solve x^2-6x-3>0

$$x ^ { 2 } - 6 x - 3 > 0$$

Solve for x

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

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.

$$x^{2}-6x-3=0$$

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, -6 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{6±4\sqrt{3}}{2}$$

Solve the equation x=\frac{6±4\sqrt{3}}{2} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

$$x-\left(2\sqrt{3}+3\right)<0 x-\left(3-2\sqrt{3}\right)<0$$

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

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

Consider the case when x-\left(2\sqrt{3}+3\right) and x-\left(3-2\sqrt{3}\right) are both positive.

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

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

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

The final solution is the union of the obtained solutions.

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