Solve -0.02x^2+168x-64,000>0

$$- 0.02 x ^ { 2 } + 168 x - 64,000 > 0$$

Solve for x

$$x\in \left(400,8000\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -0.02x^{2}+168x-64000 positive. Since -1 is negative, the inequality direction is changed.

$$0.02x^{2}-168x+64000<0$$

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.

$$0.02x^{2}-168x+64000=0$$

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 0.02 for a, -168 for b, and 64000 for c in the quadratic formula.

$$x=\frac{-\left(-168\right)±\sqrt{\left(-168\right)^{2}-4\times 0.02\times 64000}}{0.02\times 2}$$

Do the calculations.

$$x=\frac{168±152}{0.04}$$

Solve the equation x=\frac{168±152}{0.04} when ± is plus and when ± is minus.

$$x=8000 x=400$$

Rewrite the inequality by using the obtained solutions.

$$0.02\left(x-8000\right)\left(x-400\right)<0$$

For the product to be negative, x-8000 and x-400 have to be of the opposite signs. Consider the case when x-8000 is positive and x-400 is negative.

$$x-8000>0 x-400<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-400 is positive and x-8000 is negative.

$$x-400>0 x-8000<0$$

The solution satisfying both inequalities is x\in \left(400,8000\right).

$$x\in \left(400,8000\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(400,8000\right)$$