Solve +(1+5x)(100-x)=10

$$+ ( 1 + 5 x ) ( 100 - x ) = 10$$

Solve for x

$$x = \frac{\sqrt{250801} + 499}{10} \approx 99.980035942$$

$$x=\frac{499-\sqrt{250801}}{10}\approx -0.180035942$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1+5x by 100-x and combine like terms.

$$100+499x-5x^{2}=10$$

Subtract 10 from both sides.

$$100+499x-5x^{2}-10=0$$

Subtract 10 from 100 to get 90.

$$90+499x-5x^{2}=0$$

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.

$$-5x^{2}+499x+90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 499 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-499±\sqrt{499^{2}-4\left(-5\right)\times 90}}{2\left(-5\right)}$$

Square 499.

$$x=\frac{-499±\sqrt{249001-4\left(-5\right)\times 90}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-499±\sqrt{249001+20\times 90}}{2\left(-5\right)}$$

Multiply 20 times 90.

$$x=\frac{-499±\sqrt{249001+1800}}{2\left(-5\right)}$$

Add 249001 to 1800.

$$x=\frac{-499±\sqrt{250801}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-499±\sqrt{250801}}{-10}$$

Now solve the equation x=\frac{-499±\sqrt{250801}}{-10} when ± is plus. Add -499 to \sqrt{250801}.

$$x=\frac{\sqrt{250801}-499}{-10}$$

Divide -499+\sqrt{250801} by -10.

$$x=\frac{499-\sqrt{250801}}{10}$$

Now solve the equation x=\frac{-499±\sqrt{250801}}{-10} when ± is minus. Subtract \sqrt{250801} from -499.

$$x=\frac{-\sqrt{250801}-499}{-10}$$

Divide -499-\sqrt{250801} by -10.

$$x=\frac{\sqrt{250801}+499}{10}$$

The equation is now solved.

$$x=\frac{499-\sqrt{250801}}{10} x=\frac{\sqrt{250801}+499}{10}$$

Steps for Completing the Square

Use the distributive property to multiply 1+5x by 100-x and combine like terms.

$$100+499x-5x^{2}=10$$

Subtract 100 from both sides.

$$499x-5x^{2}=10-100$$

Subtract 100 from 10 to get -90.

$$499x-5x^{2}=-90$$

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.

$$-5x^{2}+499x=-90$$

Divide both sides by -5.

$$\frac{-5x^{2}+499x}{-5}=-\frac{90}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{499}{-5}x=-\frac{90}{-5}$$

Divide 499 by -5.

$$x^{2}-\frac{499}{5}x=-\frac{90}{-5}$$

Divide -90 by -5.

$$x^{2}-\frac{499}{5}x=18$$

Divide -\frac{499}{5}, the coefficient of the x term, by 2 to get -\frac{499}{10}. Then add the square of -\frac{499}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{499}{5}x+\left(-\frac{499}{10}\right)^{2}=18+\left(-\frac{499}{10}\right)^{2}$$

Square -\frac{499}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{499}{5}x+\frac{249001}{100}=18+\frac{249001}{100}$$

Add 18 to \frac{249001}{100}.

$$x^{2}-\frac{499}{5}x+\frac{249001}{100}=\frac{250801}{100}$$

Factor x^{2}-\frac{499}{5}x+\frac{249001}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{499}{10}\right)^{2}=\frac{250801}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{499}{10}\right)^{2}}=\sqrt{\frac{250801}{100}}$$

Simplify.

$$x-\frac{499}{10}=\frac{\sqrt{250801}}{10} x-\frac{499}{10}=-\frac{\sqrt{250801}}{10}$$

Add \frac{499}{10} to both sides of the equation.

$$x=\frac{\sqrt{250801}+499}{10} x=\frac{499-\sqrt{250801}}{10}$$