Solve 1000(1+x)(0.98+x)=1000(1+x)+108

$$1000(1+x)(0.98+x)=1000(1+x)+108$$

Solve for x

$$x=\frac{3\sqrt{409}-49}{100}\approx 0.116712452$$

$$x=\frac{-3\sqrt{409}-49}{100}\approx -1.096712452$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1000 by 1+x.

$$\left(1000+1000x\right)\left(0.98+x\right)=1000\left(1+x\right)+108$$

Use the distributive property to multiply 1000+1000x by 0.98+x and combine like terms.

$$980+1980x+1000x^{2}=1000\left(1+x\right)+108$$

Use the distributive property to multiply 1000 by 1+x.

$$980+1980x+1000x^{2}=1000+1000x+108$$

Add 1000 and 108 to get 1108.

$$980+1980x+1000x^{2}=1108+1000x$$

Subtract 1108 from both sides.

$$980+1980x+1000x^{2}-1108=1000x$$

Subtract 1108 from 980 to get -128.

$$-128+1980x+1000x^{2}=1000x$$

Subtract 1000x from both sides.

$$-128+1980x+1000x^{2}-1000x=0$$

Combine 1980x and -1000x to get 980x.

$$-128+980x+1000x^{2}=0$$

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.

$$1000x^{2}+980x-128=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1000 for a, 980 for b, and -128 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-980±\sqrt{980^{2}-4\times 1000\left(-128\right)}}{2\times 1000}$$

Square 980.

$$x=\frac{-980±\sqrt{960400-4\times 1000\left(-128\right)}}{2\times 1000}$$

Multiply -4 times 1000.

$$x=\frac{-980±\sqrt{960400-4000\left(-128\right)}}{2\times 1000}$$

Multiply -4000 times -128.

$$x=\frac{-980±\sqrt{960400+512000}}{2\times 1000}$$

Add 960400 to 512000.

$$x=\frac{-980±\sqrt{1472400}}{2\times 1000}$$

Take the square root of 1472400.

$$x=\frac{-980±60\sqrt{409}}{2\times 1000}$$

Multiply 2 times 1000.

$$x=\frac{-980±60\sqrt{409}}{2000}$$

Now solve the equation x=\frac{-980±60\sqrt{409}}{2000} when ± is plus. Add -980 to 60\sqrt{409}.

$$x=\frac{60\sqrt{409}-980}{2000}$$

Divide -980+60\sqrt{409} by 2000.

$$x=\frac{3\sqrt{409}-49}{100}$$

Now solve the equation x=\frac{-980±60\sqrt{409}}{2000} when ± is minus. Subtract 60\sqrt{409} from -980.

$$x=\frac{-60\sqrt{409}-980}{2000}$$

Divide -980-60\sqrt{409} by 2000.

$$x=\frac{-3\sqrt{409}-49}{100}$$

The equation is now solved.

$$x=\frac{3\sqrt{409}-49}{100} x=\frac{-3\sqrt{409}-49}{100}$$

Steps for Completing the Square

Use the distributive property to multiply 1000 by 1+x.

$$\left(1000+1000x\right)\left(0.98+x\right)=1000\left(1+x\right)+108$$

Use the distributive property to multiply 1000+1000x by 0.98+x and combine like terms.

$$980+1980x+1000x^{2}=1000\left(1+x\right)+108$$

Use the distributive property to multiply 1000 by 1+x.

$$980+1980x+1000x^{2}=1000+1000x+108$$

Add 1000 and 108 to get 1108.

$$980+1980x+1000x^{2}=1108+1000x$$

Subtract 1000x from both sides.

$$980+1980x+1000x^{2}-1000x=1108$$

Combine 1980x and -1000x to get 980x.

$$980+980x+1000x^{2}=1108$$

Subtract 980 from both sides.

$$980x+1000x^{2}=1108-980$$

Subtract 980 from 1108 to get 128.

$$980x+1000x^{2}=128$$

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.

$$1000x^{2}+980x=128$$

Divide both sides by 1000.

$$\frac{1000x^{2}+980x}{1000}=\frac{128}{1000}$$

Dividing by 1000 undoes the multiplication by 1000.

$$x^{2}+\frac{980}{1000}x=\frac{128}{1000}$$

Reduce the fraction \frac{980}{1000} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{49}{50}x=\frac{128}{1000}$$

Reduce the fraction \frac{128}{1000} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{49}{50}x=\frac{16}{125}$$

Divide \frac{49}{50}, the coefficient of the x term, by 2 to get \frac{49}{100}. Then add the square of \frac{49}{100} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{49}{50}x+\left(\frac{49}{100}\right)^{2}=\frac{16}{125}+\left(\frac{49}{100}\right)^{2}$$

Square \frac{49}{100} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{49}{50}x+\frac{2401}{10000}=\frac{16}{125}+\frac{2401}{10000}$$

Add \frac{16}{125} to \frac{2401}{10000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{49}{50}x+\frac{2401}{10000}=\frac{3681}{10000}$$

Factor x^{2}+\frac{49}{50}x+\frac{2401}{10000}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{49}{100}\right)^{2}=\frac{3681}{10000}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{49}{100}\right)^{2}}=\sqrt{\frac{3681}{10000}}$$

Simplify.

$$x+\frac{49}{100}=\frac{3\sqrt{409}}{100} x+\frac{49}{100}=-\frac{3\sqrt{409}}{100}$$

Subtract \frac{49}{100} from both sides of the equation.

$$x=\frac{3\sqrt{409}-49}{100} x=\frac{-3\sqrt{409}-49}{100}$$