Solve x-2/2-x^2-4/3leq0

$$\frac { x - 2 } { 2 } - \frac { x ^ { 2 } - 4 } { 3 } \leq 0$$

Solve for x

$$x\in (-\infty,-\frac{1}{2}]\cup [2,\infty)$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3. Since 6 is positive, the inequality direction remains the same.

$$3\left(x-2\right)-2\left(x^{2}-4\right)\leq 0$$

Use the distributive property to multiply 3 by x-2.

$$3x-6-2\left(x^{2}-4\right)\leq 0$$

Use the distributive property to multiply -2 by x^{2}-4.

$$3x-6-2x^{2}+8\leq 0$$

Add -6 and 8 to get 2.

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

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

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

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.

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

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

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

Do the calculations.

$$x=\frac{3±5}{4}$$

Solve the equation x=\frac{3±5}{4} when ± is plus and when ± is minus.

$$x=2 x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-2\right)\left(x+\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, x-2 and x+\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when x-2 and x+\frac{1}{2} are both ≤0.

$$x-2\leq 0 x+\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{1}{2}.

$$x\leq -\frac{1}{2}$$

Consider the case when x-2 and x+\frac{1}{2} are both ≥0.

$$x+\frac{1}{2}\geq 0 x-2\geq 0$$

The solution satisfying both inequalities is x\geq 2.

$$x\geq 2$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{1}{2}\text{; }x\geq 2$$