Solve (10-x)(200+50x)=1250

$$(10-x)(200+50x)=1250$$

Solve for x

$$x=2\sqrt{6}+3\approx 7.898979486$$

$$x=3-2\sqrt{6}\approx -1.898979486$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 10-x by 200+50x and combine like terms.

$$2000+300x-50x^{2}=1250$$

Subtract 1250 from both sides.

$$2000+300x-50x^{2}-1250=0$$

Subtract 1250 from 2000 to get 750.

$$750+300x-50x^{2}=0$$

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.

$$-50x^{2}+300x+750=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -50 for a, 300 for b, and 750 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-300±\sqrt{300^{2}-4\left(-50\right)\times 750}}{2\left(-50\right)}$$

Square 300.

$$x=\frac{-300±\sqrt{90000-4\left(-50\right)\times 750}}{2\left(-50\right)}$$

Multiply -4 times -50.

$$x=\frac{-300±\sqrt{90000+200\times 750}}{2\left(-50\right)}$$

Multiply 200 times 750.

$$x=\frac{-300±\sqrt{90000+150000}}{2\left(-50\right)}$$

Add 90000 to 150000.

$$x=\frac{-300±\sqrt{240000}}{2\left(-50\right)}$$

Take the square root of 240000.

$$x=\frac{-300±200\sqrt{6}}{2\left(-50\right)}$$

Multiply 2 times -50.

$$x=\frac{-300±200\sqrt{6}}{-100}$$

Now solve the equation x=\frac{-300±200\sqrt{6}}{-100} when ± is plus. Add -300 to 200\sqrt{6}.

$$x=\frac{200\sqrt{6}-300}{-100}$$

Divide -300+200\sqrt{6} by -100.

$$x=3-2\sqrt{6}$$

Now solve the equation x=\frac{-300±200\sqrt{6}}{-100} when ± is minus. Subtract 200\sqrt{6} from -300.

$$x=\frac{-200\sqrt{6}-300}{-100}$$

Divide -300-200\sqrt{6} by -100.

$$x=2\sqrt{6}+3$$

The equation is now solved.

$$x=3-2\sqrt{6} x=2\sqrt{6}+3$$

Steps for Completing the Square

Use the distributive property to multiply 10-x by 200+50x and combine like terms.

$$2000+300x-50x^{2}=1250$$

Subtract 2000 from both sides.

$$300x-50x^{2}=1250-2000$$

Subtract 2000 from 1250 to get -750.

$$300x-50x^{2}=-750$$

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.

$$-50x^{2}+300x=-750$$

Divide both sides by -50.

$$\frac{-50x^{2}+300x}{-50}=-\frac{750}{-50}$$

Dividing by -50 undoes the multiplication by -50.

$$x^{2}+\frac{300}{-50}x=-\frac{750}{-50}$$

Divide 300 by -50.

$$x^{2}-6x=-\frac{750}{-50}$$

Divide -750 by -50.

$$x^{2}-6x=15$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=15+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=15+9$$

Add 15 to 9.

$$x^{2}-6x+9=24$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{24}$$

Simplify.

$$x-3=2\sqrt{6} x-3=-2\sqrt{6}$$

Add 3 to both sides of the equation.

$$x=2\sqrt{6}+3 x=3-2\sqrt{6}$$