Solve 5r^2-16=16r

$$5 r ^ { 2 } - 16 = 16 r$$

Solve for r

$$r=-\frac{4}{5}=-0.8$$

$$r=4$$

Steps Using Factoring By Grouping

Subtract 16r from both sides.

$$5r^{2}-16-16r=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5r^{2}-16r-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5r^{2}+ar+br-16. To find a and b, set up a system to be solved.

$$a+b=-16 ab=5\left(-16\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -16.

$$a=-20 b=4$$

Rewrite 5r^{2}-16r-16 as \left(5r^{2}-20r\right)+\left(4r-16\right).

$$\left(5r^{2}-20r\right)+\left(4r-16\right)$$

Factor out 5r in the first and 4 in the second group.

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

Factor out common term r-4 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 16r from both sides.

$$5r^{2}-16-16r=0$$

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.

$$5r^{2}-16r-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -16 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\left(-16\right)}}{2\times 5}$$

Square -16.

$$r=\frac{-\left(-16\right)±\sqrt{256-4\times 5\left(-16\right)}}{2\times 5}$$

Multiply -4 times 5.

$$r=\frac{-\left(-16\right)±\sqrt{256-20\left(-16\right)}}{2\times 5}$$

Multiply -20 times -16.

$$r=\frac{-\left(-16\right)±\sqrt{256+320}}{2\times 5}$$

Add 256 to 320.

$$r=\frac{-\left(-16\right)±\sqrt{576}}{2\times 5}$$

Take the square root of 576.

$$r=\frac{-\left(-16\right)±24}{2\times 5}$$

The opposite of -16 is 16.

$$r=\frac{16±24}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{16±24}{10}$$

Now solve the equation r=\frac{16±24}{10} when ± is plus. Add 16 to 24.

$$r=\frac{40}{10}$$

Divide 40 by 10.

$$r=4$$

Now solve the equation r=\frac{16±24}{10} when ± is minus. Subtract 24 from 16.

$$r=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 16r from both sides.

$$5r^{2}-16-16r=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$5r^{2}-16r=16$$

Divide both sides by 5.

$$\frac{5r^{2}-16r}{5}=\frac{16}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$r^{2}-\frac{16}{5}r=\frac{16}{5}$$

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{16}{5}r+\left(-\frac{8}{5}\right)^{2}=\frac{16}{5}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{16}{5}r+\frac{64}{25}=\frac{16}{5}+\frac{64}{25}$$

Add \frac{16}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{16}{5}r+\frac{64}{25}=\frac{144}{25}$$

Factor r^{2}-\frac{16}{5}r+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{8}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{8}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$r-\frac{8}{5}=\frac{12}{5} r-\frac{8}{5}=-\frac{12}{5}$$

Add \frac{8}{5} to both sides of the equation.

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