Solve -5x^2-24x+36leq0

$$- 5 x ^ { 2 } - 24 x + 36 \leq 0$$

Solve for x

$$x\in (-\infty,-6]\cup [\frac{6}{5},\infty)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -5x^{2}-24x+36 positive. Since -1 is negative, the inequality direction is changed.

$$5x^{2}+24x-36\geq 0$$

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.

$$5x^{2}+24x-36=0$$

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 5 for a, 24 for b, and -36 for c in the quadratic formula.

$$x=\frac{-24±\sqrt{24^{2}-4\times 5\left(-36\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{-24±36}{10}$$

Solve the equation x=\frac{-24±36}{10} when ± is plus and when ± is minus.

$$x=\frac{6}{5} x=-6$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-\frac{6}{5}\right)\left(x+6\right)\geq 0$$

For the product to be ≥0, x-\frac{6}{5} and x+6 have to be both ≤0 or both ≥0. Consider the case when x-\frac{6}{5} and x+6 are both ≤0.

$$x-\frac{6}{5}\leq 0 x+6\leq 0$$

The solution satisfying both inequalities is x\leq -6.

$$x\leq -6$$

Consider the case when x-\frac{6}{5} and x+6 are both ≥0.

$$x+6\geq 0 x-\frac{6}{5}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{6}{5}.

$$x\geq \frac{6}{5}$$

The final solution is the union of the obtained solutions.

$$x\leq -6\text{; }x\geq \frac{6}{5}$$