Solve -100+601/1+r+721/left(1+rright)^2=0

$$-100+60 \frac{ 1 }{ 1+r } +72 \frac{ 1 }{ { \left(1+r \right) }^{ 2 } } = 0$$

Solve for r

$$r=\frac{1}{5}=0.2$$

$$r = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

Steps Using Factoring By Grouping

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}, the least common multiple of 1+r,\left(1+r\right)^{2}.

$$\left(r+1\right)^{2}\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(r+1\right)^{2}.

$$\left(r^{2}+2r+1\right)\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use the distributive property to multiply r^{2}+2r+1 by -100.

$$-100r^{2}-200r-100+60\left(r+1\right)\times 1+72\times 1=0$$

Multiply 60 and 1 to get 60.

$$-100r^{2}-200r-100+60\left(r+1\right)+72\times 1=0$$

Use the distributive property to multiply 60 by r+1.

$$-100r^{2}-200r-100+60r+60+72\times 1=0$$

Combine -200r and 60r to get -140r.

$$-100r^{2}-140r-100+60+72\times 1=0$$

Add -100 and 60 to get -40.

$$-100r^{2}-140r-40+72\times 1=0$$

Multiply 72 and 1 to get 72.

$$-100r^{2}-140r-40+72=0$$

Add -40 and 72 to get 32.

$$-100r^{2}-140r+32=0$$

Divide both sides by 4.

$$-25r^{2}-35r+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -25r^{2}+ar+br+8. To find a and b, set up a system to be solved.

$$a+b=-35 ab=-25\times 8=-200$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -200.

$$1,-200 2,-100 4,-50 5,-40 8,-25 10,-20$$

Calculate the sum for each pair.

$$1-200=-199 2-100=-98 4-50=-46 5-40=-35 8-25=-17 10-20=-10$$

The solution is the pair that gives sum -35.

$$a=5 b=-40$$

Rewrite -25r^{2}-35r+8 as \left(-25r^{2}+5r\right)+\left(-40r+8\right).

$$\left(-25r^{2}+5r\right)+\left(-40r+8\right)$$

Factor out 5r in the first and 8 in the second group.

$$5r\left(-5r+1\right)+8\left(-5r+1\right)$$

Factor out common term -5r+1 by using distributive property.

$$\left(-5r+1\right)\left(5r+8\right)$$

To find equation solutions, solve -5r+1=0 and 5r+8=0.

$$r=\frac{1}{5} r=-\frac{8}{5}$$

Steps Using the Quadratic Formula

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}, the least common multiple of 1+r,\left(1+r\right)^{2}.

$$\left(r+1\right)^{2}\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(r+1\right)^{2}.

$$\left(r^{2}+2r+1\right)\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use the distributive property to multiply r^{2}+2r+1 by -100.

$$-100r^{2}-200r-100+60\left(r+1\right)\times 1+72\times 1=0$$

Multiply 60 and 1 to get 60.

$$-100r^{2}-200r-100+60\left(r+1\right)+72\times 1=0$$

Use the distributive property to multiply 60 by r+1.

$$-100r^{2}-200r-100+60r+60+72\times 1=0$$

Combine -200r and 60r to get -140r.

$$-100r^{2}-140r-100+60+72\times 1=0$$

Add -100 and 60 to get -40.

$$-100r^{2}-140r-40+72\times 1=0$$

Multiply 72 and 1 to get 72.

$$-100r^{2}-140r-40+72=0$$

Add -40 and 72 to get 32.

$$-100r^{2}-140r+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -100 for a, -140 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-140\right)±\sqrt{\left(-140\right)^{2}-4\left(-100\right)\times 32}}{2\left(-100\right)}$$

Square -140.

$$r=\frac{-\left(-140\right)±\sqrt{19600-4\left(-100\right)\times 32}}{2\left(-100\right)}$$

Multiply -4 times -100.

$$r=\frac{-\left(-140\right)±\sqrt{19600+400\times 32}}{2\left(-100\right)}$$

Multiply 400 times 32.

$$r=\frac{-\left(-140\right)±\sqrt{19600+12800}}{2\left(-100\right)}$$

Add 19600 to 12800.

$$r=\frac{-\left(-140\right)±\sqrt{32400}}{2\left(-100\right)}$$

Take the square root of 32400.

$$r=\frac{-\left(-140\right)±180}{2\left(-100\right)}$$

The opposite of -140 is 140.

$$r=\frac{140±180}{2\left(-100\right)}$$

Multiply 2 times -100.

$$r=\frac{140±180}{-200}$$

Now solve the equation r=\frac{140±180}{-200} when ± is plus. Add 140 to 180.

$$r=\frac{320}{-200}$$

Reduce the fraction \frac{320}{-200} to lowest terms by extracting and canceling out 40.

$$r=-\frac{8}{5}$$

Now solve the equation r=\frac{140±180}{-200} when ± is minus. Subtract 180 from 140.

$$r=-\frac{40}{-200}$$

Reduce the fraction \frac{-40}{-200} to lowest terms by extracting and canceling out 40.

$$r=\frac{1}{5}$$

The equation is now solved.

$$r=-\frac{8}{5} r=\frac{1}{5}$$

Steps for Completing the Square

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}, the least common multiple of 1+r,\left(1+r\right)^{2}.

$$\left(r+1\right)^{2}\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(r+1\right)^{2}.

$$\left(r^{2}+2r+1\right)\left(-100\right)+60\left(r+1\right)\times 1+72\times 1=0$$

Use the distributive property to multiply r^{2}+2r+1 by -100.

$$-100r^{2}-200r-100+60\left(r+1\right)\times 1+72\times 1=0$$

Multiply 60 and 1 to get 60.

$$-100r^{2}-200r-100+60\left(r+1\right)+72\times 1=0$$

Use the distributive property to multiply 60 by r+1.

$$-100r^{2}-200r-100+60r+60+72\times 1=0$$

Combine -200r and 60r to get -140r.

$$-100r^{2}-140r-100+60+72\times 1=0$$

Add -100 and 60 to get -40.

$$-100r^{2}-140r-40+72\times 1=0$$

Multiply 72 and 1 to get 72.

$$-100r^{2}-140r-40+72=0$$

Add -40 and 72 to get 32.

$$-100r^{2}-140r+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-100r^{2}-140r=-32$$

Divide both sides by -100.

$$\frac{-100r^{2}-140r}{-100}=-\frac{32}{-100}$$

Dividing by -100 undoes the multiplication by -100.

$$r^{2}+\left(-\frac{140}{-100}\right)r=-\frac{32}{-100}$$

Reduce the fraction \frac{-140}{-100} to lowest terms by extracting and canceling out 20.

$$r^{2}+\frac{7}{5}r=-\frac{32}{-100}$$

Reduce the fraction \frac{-32}{-100} to lowest terms by extracting and canceling out 4.

$$r^{2}+\frac{7}{5}r=\frac{8}{25}$$

Divide \frac{7}{5}, the coefficient of the x term, by 2 to get \frac{7}{10}. Then add the square of \frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+\frac{7}{5}r+\left(\frac{7}{10}\right)^{2}=\frac{8}{25}+\left(\frac{7}{10}\right)^{2}$$

Square \frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$r^{2}+\frac{7}{5}r+\frac{49}{100}=\frac{8}{25}+\frac{49}{100}$$

Add \frac{8}{25} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}+\frac{7}{5}r+\frac{49}{100}=\frac{81}{100}$$

Factor r^{2}+\frac{7}{5}r+\frac{49}{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}.

$$\left(r+\frac{7}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{7}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$r+\frac{7}{10}=\frac{9}{10} r+\frac{7}{10}=-\frac{9}{10}$$

Subtract \frac{7}{10} from both sides of the equation.

$$r=\frac{1}{5} r=-\frac{8}{5}$$