Solve -1/2x^2+20x-500leq0

$$- \frac { 1 } { 2 } x ^ { 2 } + 20 x - 500 \leq 0$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -\frac{1}{2}x^{2}+20x-500 positive. Since -1 is negative, the inequality direction is changed.

$$\frac{1}{2}x^{2}-20x+500\geq 0$$

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.

$$\frac{1}{2}x^{2}-20x+500=0$$

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 \frac{1}{2} for a, -20 for b, and 500 for c in the quadratic formula.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times \frac{1}{2}\times 500}}{\frac{1}{2}\times 2}$$

Do the calculations.

$$x=\frac{20±\sqrt{-600}}{1}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression \frac{1}{2}x^{2}-20x+500 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$\frac{1}{2}\times 0^{2}-20\times 0+500=500$$

The value of the expression \frac{1}{2}x^{2}-20x+500 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$