Solve -5x^2+700x-4420-20000geq0

$$-5 { x }^{ 2 } +700x-4420-20000 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}66,74\end{bmatrix}$$

Steps Using the Quadratic Formula

Subtract 20000 from -4420 to get -24420.

$$-5x^{2}+700x-24420\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -5x^{2}+700x-24420 positive. Since -1 is negative, the inequality direction is changed.

$$5x^{2}-700x+24420\leq 0$$

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.

$$5x^{2}-700x+24420=0$$

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, -700 for b, and 24420 for c in the quadratic formula.

$$x=\frac{-\left(-700\right)±\sqrt{\left(-700\right)^{2}-4\times 5\times 24420}}{2\times 5}$$

Do the calculations.

$$x=\frac{700±40}{10}$$

Solve the equation x=\frac{700±40}{10} when ± is plus and when ± is minus.

$$x=74 x=66$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-74\right)\left(x-66\right)\leq 0$$

For the product to be ≤0, one of the values x-74 and x-66 has to be ≥0 and the other has to be ≤0. Consider the case when x-74\geq 0 and x-66\leq 0.

$$x-74\geq 0 x-66\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-74\leq 0 and x-66\geq 0.

$$x-66\geq 0 x-74\leq 0$$

The solution satisfying both inequalities is x\in \left[66,74\right].

$$x\in \begin{bmatrix}66,74\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}66,74\end{bmatrix}$$