Solve -x^2+2x+3leq0

$$- x ^ { 2 } + 2 x + 3 \leq 0$$

Solve for x

$$x\in (-\infty,-1]\cup [3,\infty)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+2x+3 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-2x-3\geq 0$$

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.

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

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, -2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±4}{2}$$

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

$$x=3 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x+1\right)\geq 0$$

For the product to be ≥0, x-3 and x+1 have to be both ≤0 or both ≥0. Consider the case when x-3 and x+1 are both ≤0.

$$x-3\leq 0 x+1\leq 0$$

The solution satisfying both inequalities is x\leq -1.

$$x\leq -1$$

Consider the case when x-3 and x+1 are both ≥0.

$$x+1\geq 0 x-3\geq 0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq -1\text{; }x\geq 3$$