Solve 1250=1/10x(300-x)

$$1250= \frac{ 1 }{ 10 } x(300-x)$$

Solve for x

$$x=50$$

$$x=250$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{10}x by 300-x.

$$1250=\frac{1}{10}x\times 300+\frac{1}{10}x\left(-1\right)x$$

Multiply x and x to get x^{2}.

$$1250=\frac{1}{10}x\times 300+\frac{1}{10}x^{2}\left(-1\right)$$

Multiply \frac{1}{10} and 300 to get \frac{300}{10}.

$$1250=\frac{300}{10}x+\frac{1}{10}x^{2}\left(-1\right)$$

Divide 300 by 10 to get 30.

$$1250=30x+\frac{1}{10}x^{2}\left(-1\right)$$

Multiply \frac{1}{10} and -1 to get -\frac{1}{10}.

$$1250=30x-\frac{1}{10}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$30x-\frac{1}{10}x^{2}=1250$$

Subtract 1250 from both sides.

$$30x-\frac{1}{10}x^{2}-1250=0$$

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.

$$-\frac{1}{10}x^{2}+30x-1250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{10} for a, 30 for b, and -1250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-\frac{1}{10}\right)\left(-1250\right)}}{2\left(-\frac{1}{10}\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-\frac{1}{10}\right)\left(-1250\right)}}{2\left(-\frac{1}{10}\right)}$$

Multiply -4 times -\frac{1}{10}.

$$x=\frac{-30±\sqrt{900+\frac{2}{5}\left(-1250\right)}}{2\left(-\frac{1}{10}\right)}$$

Multiply \frac{2}{5} times -1250.

$$x=\frac{-30±\sqrt{900-500}}{2\left(-\frac{1}{10}\right)}$$

Add 900 to -500.

$$x=\frac{-30±\sqrt{400}}{2\left(-\frac{1}{10}\right)}$$

Take the square root of 400.

$$x=\frac{-30±20}{2\left(-\frac{1}{10}\right)}$$

Multiply 2 times -\frac{1}{10}.

$$x=\frac{-30±20}{-\frac{1}{5}}$$

Now solve the equation x=\frac{-30±20}{-\frac{1}{5}} when ± is plus. Add -30 to 20.

$$x=-\frac{10}{-\frac{1}{5}}$$

Divide -10 by -\frac{1}{5} by multiplying -10 by the reciprocal of -\frac{1}{5}.

$$x=50$$

Now solve the equation x=\frac{-30±20}{-\frac{1}{5}} when ± is minus. Subtract 20 from -30.

$$x=-\frac{50}{-\frac{1}{5}}$$

Divide -50 by -\frac{1}{5} by multiplying -50 by the reciprocal of -\frac{1}{5}.

$$x=250$$

The equation is now solved.

$$x=50 x=250$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{10}x by 300-x.

$$1250=\frac{1}{10}x\times 300+\frac{1}{10}x\left(-1\right)x$$

Multiply x and x to get x^{2}.

$$1250=\frac{1}{10}x\times 300+\frac{1}{10}x^{2}\left(-1\right)$$

Multiply \frac{1}{10} and 300 to get \frac{300}{10}.

$$1250=\frac{300}{10}x+\frac{1}{10}x^{2}\left(-1\right)$$

Divide 300 by 10 to get 30.

$$1250=30x+\frac{1}{10}x^{2}\left(-1\right)$$

Multiply \frac{1}{10} and -1 to get -\frac{1}{10}.

$$1250=30x-\frac{1}{10}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$30x-\frac{1}{10}x^{2}=1250$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-\frac{1}{10}x^{2}+30x=1250$$

Multiply both sides by -10.

$$\frac{-\frac{1}{10}x^{2}+30x}{-\frac{1}{10}}=\frac{1250}{-\frac{1}{10}}$$

Dividing by -\frac{1}{10} undoes the multiplication by -\frac{1}{10}.

$$x^{2}+\frac{30}{-\frac{1}{10}}x=\frac{1250}{-\frac{1}{10}}$$

Divide 30 by -\frac{1}{10} by multiplying 30 by the reciprocal of -\frac{1}{10}.

$$x^{2}-300x=\frac{1250}{-\frac{1}{10}}$$

Divide 1250 by -\frac{1}{10} by multiplying 1250 by the reciprocal of -\frac{1}{10}.

$$x^{2}-300x=-12500$$

Divide -300, the coefficient of the x term, by 2 to get -150. Then add the square of -150 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-300x+\left(-150\right)^{2}=-12500+\left(-150\right)^{2}$$

Square -150.

$$x^{2}-300x+22500=-12500+22500$$

Add -12500 to 22500.

$$x^{2}-300x+22500=10000$$

Factor x^{2}-300x+22500. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-150\right)^{2}=10000$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-150\right)^{2}}=\sqrt{10000}$$

Simplify.

$$x-150=100 x-150=-100$$

Add 150 to both sides of the equation.

$$x=250 x=50$$