Solve x^2+6x-3=geq0

$$x ^ { 2 } + 6 x - 3 = \geq 0$$

Solve for x

$$x\in (-\infty,-2\sqrt{3}-3]\cup [2\sqrt{3}-3,\infty)$$

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.

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

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.

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

Rewrite the inequality by using the obtained solutions.

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

For the product to be ≥0, x-\left(2\sqrt{3}-3\right) and x-\left(-2\sqrt{3}-3\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(2\sqrt{3}-3\right) and x-\left(-2\sqrt{3}-3\right) are both ≤0.

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

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

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

Consider the case when x-\left(2\sqrt{3}-3\right) and x-\left(-2\sqrt{3}-3\right) are both ≥0.

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

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

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

The final solution is the union of the obtained solutions.

$$x\leq -2\sqrt{3}-3\text{; }x\geq 2\sqrt{3}-3$$