Solve x^2+36x-220geq0

$$x ^ { 2 } + 36 x - 220 \geq 0$$

Solve for x

$$x\in (-\infty,-4\sqrt{34}-18]\cup [4\sqrt{34}-18,\infty)$$

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.

$$x^{2}+36x-220=0$$

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, 36 for b, and -220 for c in the quadratic formula.

$$x=\frac{-36±\sqrt{36^{2}-4\times 1\left(-220\right)}}{2}$$

Do the calculations.

$$x=\frac{-36±8\sqrt{34}}{2}$$

Solve the equation x=\frac{-36±8\sqrt{34}}{2} when ± is plus and when ± is minus.

$$x=4\sqrt{34}-18 x=-4\sqrt{34}-18$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(4\sqrt{34}-18\right)\right)\left(x-\left(-4\sqrt{34}-18\right)\right)\geq 0$$

For the product to be ≥0, x-\left(4\sqrt{34}-18\right) and x-\left(-4\sqrt{34}-18\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(4\sqrt{34}-18\right) and x-\left(-4\sqrt{34}-18\right) are both ≤0.

$$x-\left(4\sqrt{34}-18\right)\leq 0 x-\left(-4\sqrt{34}-18\right)\leq 0$$

The solution satisfying both inequalities is x\leq -4\sqrt{34}-18.

$$x\leq -4\sqrt{34}-18$$

Consider the case when x-\left(4\sqrt{34}-18\right) and x-\left(-4\sqrt{34}-18\right) are both ≥0.

$$x-\left(-4\sqrt{34}-18\right)\geq 0 x-\left(4\sqrt{34}-18\right)\geq 0$$

The solution satisfying both inequalities is x\geq 4\sqrt{34}-18.

$$x\geq 4\sqrt{34}-18$$

The final solution is the union of the obtained solutions.

$$x\leq -4\sqrt{34}-18\text{; }x\geq 4\sqrt{34}-18$$