Solve 9r(r-2)=r^2+5

$$9 r ( r - 2 ) = r ^ { 2 } + 5$$

Solve for r

$$r=-\frac{1}{4}=-0.25$$

$$r = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 9r by r-2.

$$9r^{2}-18r=r^{2}+5$$

Subtract r^{2} from both sides.

$$9r^{2}-18r-r^{2}=5$$

Combine 9r^{2} and -r^{2} to get 8r^{2}.

$$8r^{2}-18r=5$$

Subtract 5 from both sides.

$$8r^{2}-18r-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8r^{2}+ar+br-5. To find a and b, set up a system to be solved.

$$a+b=-18 ab=8\left(-5\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -18.

$$a=-20 b=2$$

Rewrite 8r^{2}-18r-5 as \left(8r^{2}-20r\right)+\left(2r-5\right).

$$\left(8r^{2}-20r\right)+\left(2r-5\right)$$

Factor out 4r in 8r^{2}-20r.

$$4r\left(2r-5\right)+2r-5$$

Factor out common term 2r-5 by using distributive property.

$$\left(2r-5\right)\left(4r+1\right)$$

To find equation solutions, solve 2r-5=0 and 4r+1=0.

$$r=\frac{5}{2} r=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9r by r-2.

$$9r^{2}-18r=r^{2}+5$$

Subtract r^{2} from both sides.

$$9r^{2}-18r-r^{2}=5$$

Combine 9r^{2} and -r^{2} to get 8r^{2}.

$$8r^{2}-18r=5$$

Subtract 5 from both sides.

$$8r^{2}-18r-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -18 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 8\left(-5\right)}}{2\times 8}$$

Square -18.

$$r=\frac{-\left(-18\right)±\sqrt{324-4\times 8\left(-5\right)}}{2\times 8}$$

Multiply -4 times 8.

$$r=\frac{-\left(-18\right)±\sqrt{324-32\left(-5\right)}}{2\times 8}$$

Multiply -32 times -5.

$$r=\frac{-\left(-18\right)±\sqrt{324+160}}{2\times 8}$$

Add 324 to 160.

$$r=\frac{-\left(-18\right)±\sqrt{484}}{2\times 8}$$

Take the square root of 484.

$$r=\frac{-\left(-18\right)±22}{2\times 8}$$

The opposite of -18 is 18.

$$r=\frac{18±22}{2\times 8}$$

Multiply 2 times 8.

$$r=\frac{18±22}{16}$$

Now solve the equation r=\frac{18±22}{16} when ± is plus. Add 18 to 22.

$$r=\frac{40}{16}$$

Reduce the fraction \frac{40}{16} to lowest terms by extracting and canceling out 8.

$$r=\frac{5}{2}$$

Now solve the equation r=\frac{18±22}{16} when ± is minus. Subtract 22 from 18.

$$r=-\frac{4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$r=-\frac{1}{4}$$

The equation is now solved.

$$r=\frac{5}{2} r=-\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 9r by r-2.

$$9r^{2}-18r=r^{2}+5$$

Subtract r^{2} from both sides.

$$9r^{2}-18r-r^{2}=5$$

Combine 9r^{2} and -r^{2} to get 8r^{2}.

$$8r^{2}-18r=5$$

Divide both sides by 8.

$$\frac{8r^{2}-18r}{8}=\frac{5}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$r^{2}+\left(-\frac{18}{8}\right)r=\frac{5}{8}$$

Reduce the fraction \frac{-18}{8} to lowest terms by extracting and canceling out 2.

$$r^{2}-\frac{9}{4}r=\frac{5}{8}$$

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{9}{4}r+\left(-\frac{9}{8}\right)^{2}=\frac{5}{8}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{9}{4}r+\frac{81}{64}=\frac{5}{8}+\frac{81}{64}$$

Add \frac{5}{8} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{9}{4}r+\frac{81}{64}=\frac{121}{64}$$

Factor r^{2}-\frac{9}{4}r+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{9}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{9}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$r-\frac{9}{8}=\frac{11}{8} r-\frac{9}{8}=-\frac{11}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$r=\frac{5}{2} r=-\frac{1}{4}$$