Solve 500x-10x^2

$$500x-10 { x }^{ 2 }$$

Factor

$$10x\left(50-x\right)$$

Solution Steps

Factor out 10.

$$10\left(50x-x^{2}\right)$$

Consider 50x-x^{2}. Factor out x.

$$x\left(50-x\right)$$

Rewrite the complete factored expression.

$$10x\left(-x+50\right)$$

Steps Using the Quadratic Formula

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.

$$-10x^{2}+500x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-500±\sqrt{500^{2}}}{2\left(-10\right)}$$

Take the square root of 500^{2}.

$$x=\frac{-500±500}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-500±500}{-20}$$

Now solve the equation x=\frac{-500±500}{-20} when ± is plus. Add -500 to 500.

$$x=\frac{0}{-20}$$

Divide 0 by -20.

$$x=0$$

Now solve the equation x=\frac{-500±500}{-20} when ± is minus. Subtract 500 from -500.

$$x=-\frac{1000}{-20}$$

Divide -1000 by -20.

$$x=50$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 50 for x_{2}.

$$-10x^{2}+500x=-10x\left(x-50\right)$$

Evaluate

$$10x\left(50-x\right)$$