Solve 80x-25x^2-36leq0

$$80 x - 25 x ^ { 2 } - 36 \leq 0$$

Solve for x

$$x\in (-\infty,\frac{8-2\sqrt{7}}{5}]\cup [\frac{2\sqrt{7}+8}{5},\infty)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 80x-25x^{2}-36 positive. Since -1 is negative, the inequality direction is changed.

$$-80x+25x^{2}+36\geq 0$$

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.

$$-80x+25x^{2}+36=0$$

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 25 for a, -80 for b, and 36 for c in the quadratic formula.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 25\times 36}}{2\times 25}$$

Do the calculations.

$$x=\frac{80±20\sqrt{7}}{50}$$

Solve the equation x=\frac{80±20\sqrt{7}}{50} when ± is plus and when ± is minus.

$$x=\frac{2\sqrt{7}+8}{5} x=\frac{8-2\sqrt{7}}{5}$$

Rewrite the inequality by using the obtained solutions.

$$25\left(x-\frac{2\sqrt{7}+8}{5}\right)\left(x-\frac{8-2\sqrt{7}}{5}\right)\geq 0$$

For the product to be ≥0, x-\frac{2\sqrt{7}+8}{5} and x-\frac{8-2\sqrt{7}}{5} have to be both ≤0 or both ≥0. Consider the case when x-\frac{2\sqrt{7}+8}{5} and x-\frac{8-2\sqrt{7}}{5} are both ≤0.

$$x-\frac{2\sqrt{7}+8}{5}\leq 0 x-\frac{8-2\sqrt{7}}{5}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{8-2\sqrt{7}}{5}.

$$x\leq \frac{8-2\sqrt{7}}{5}$$

Consider the case when x-\frac{2\sqrt{7}+8}{5} and x-\frac{8-2\sqrt{7}}{5} are both ≥0.

$$x-\frac{8-2\sqrt{7}}{5}\geq 0 x-\frac{2\sqrt{7}+8}{5}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{2\sqrt{7}+8}{5}.

$$x\geq \frac{2\sqrt{7}+8}{5}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{8-2\sqrt{7}}{5}\text{; }x\geq \frac{2\sqrt{7}+8}{5}$$