Solve x^2+2x(1-15)+5geq0

$$x ^ { 2 } + 2 x ( 1 - 15 ) + 5 \geq 0$$

Solve for x

$$x\in (-\infty,14-\sqrt{191}]\cup [\sqrt{191}+14,\infty)$$

Steps Using the Quadratic Formula

Subtract 15 from 1 to get -14.

$$x^{2}+2x\left(-14\right)+5\geq 0$$

Multiply 2 and -14 to get -28.

$$x^{2}-28x+5\geq 0$$

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.

$$x^{2}-28x+5=0$$

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, -28 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{28±2\sqrt{191}}{2}$$

Solve the equation x=\frac{28±2\sqrt{191}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{191}+14 x=14-\sqrt{191}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{191}+14\right)\right)\left(x-\left(14-\sqrt{191}\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\sqrt{191}+14\right) and x-\left(14-\sqrt{191}\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\sqrt{191}+14\right) and x-\left(14-\sqrt{191}\right) are both ≤0.

$$x-\left(\sqrt{191}+14\right)\leq 0 x-\left(14-\sqrt{191}\right)\leq 0$$

The solution satisfying both inequalities is x\leq 14-\sqrt{191}.

$$x\leq 14-\sqrt{191}$$

Consider the case when x-\left(\sqrt{191}+14\right) and x-\left(14-\sqrt{191}\right) are both ≥0.

$$x-\left(14-\sqrt{191}\right)\geq 0 x-\left(\sqrt{191}+14\right)\geq 0$$

The solution satisfying both inequalities is x\geq \sqrt{191}+14.

$$x\geq \sqrt{191}+14$$

The final solution is the union of the obtained solutions.

$$x\leq 14-\sqrt{191}\text{; }x\geq \sqrt{191}+14$$