Solve (50-(x-100/5))x-1100>0

$$(50-( \frac{ x-100 }{ 5 } ))x-1100 > 0$$

Solve for x

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5. Since 5 is positive, the inequality direction remains the same.

$$5\left(50-\frac{x-100}{5}\right)x-5500>0$$

Use the distributive property to multiply 5 by 50-\frac{x-100}{5}.

$$\left(250+5\left(-\frac{x-100}{5}\right)\right)x-5500>0$$

Express 5\left(-\frac{x-100}{5}\right) as a single fraction.

$$\left(250+\frac{-5\left(x-100\right)}{5}\right)x-5500>0$$

Cancel out 5 and 5.

$$\left(250-\left(x-100\right)\right)x-5500>0$$

To find the opposite of x-100, find the opposite of each term.

$$\left(250-x-\left(-100\right)\right)x-5500>0$$

The opposite of -100 is 100.

$$\left(250-x+100\right)x-5500>0$$

Add 250 and 100 to get 350.

$$\left(350-x\right)x-5500>0$$

Use the distributive property to multiply 350-x by x.

$$350x-x^{2}-5500>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 350x-x^{2}-5500 positive. Since -1 is negative, the inequality direction is changed.

$$-350x+x^{2}+5500<0$$

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.

$$-350x+x^{2}+5500=0$$

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 1 for a, -350 for b, and 5500 for c in the quadratic formula.

$$x=\frac{-\left(-350\right)±\sqrt{\left(-350\right)^{2}-4\times 1\times 5500}}{2}$$

Do the calculations.

$$x=\frac{350±10\sqrt{1005}}{2}$$

Solve the equation x=\frac{350±10\sqrt{1005}}{2} when ± is plus and when ± is minus.

$$x=5\sqrt{1005}+175 x=175-5\sqrt{1005}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(5\sqrt{1005}+175\right)\right)\left(x-\left(175-5\sqrt{1005}\right)\right)<0$$

For the product to be negative, x-\left(5\sqrt{1005}+175\right) and x-\left(175-5\sqrt{1005}\right) have to be of the opposite signs. Consider the case when x-\left(5\sqrt{1005}+175\right) is positive and x-\left(175-5\sqrt{1005}\right) is negative.

$$x-\left(5\sqrt{1005}+175\right)>0 x-\left(175-5\sqrt{1005}\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(175-5\sqrt{1005}\right) is positive and x-\left(5\sqrt{1005}+175\right) is negative.

$$x-\left(175-5\sqrt{1005}\right)>0 x-\left(5\sqrt{1005}+175\right)<0$$

The solution satisfying both inequalities is x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right).

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$