Solve r^2=32-4r

$$r ^ { 2 } = 32 - 4 r$$

Solve for r

$$r=-8$$

$$r=4$$

Steps Using Factoring

Subtract 32 from both sides.

$$r^{2}-32=-4r$$

Add 4r to both sides.

$$r^{2}-32+4r=0$$

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

$$r^{2}+4r-32=0$$

To solve the equation, factor r^{2}+4r-32 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=-4 b=8$$

Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.

$$\left(r-4\right)\left(r+8\right)$$

To find equation solutions, solve r-4=0 and r+8=0.

$$r=4 r=-8$$

Steps Using Factoring By Grouping

Subtract 32 from both sides.

$$r^{2}-32=-4r$$

Add 4r to both sides.

$$r^{2}-32+4r=0$$

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

$$r^{2}+4r-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-32. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-32\right)=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=-4 b=8$$

Rewrite r^{2}+4r-32 as \left(r^{2}-4r\right)+\left(8r-32\right).

$$\left(r^{2}-4r\right)+\left(8r-32\right)$$

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

$$r\left(r-4\right)+8\left(r-4\right)$$

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

$$\left(r-4\right)\left(r+8\right)$$

To find equation solutions, solve r-4=0 and r+8=0.

$$r=4 r=-8$$

Steps Using the Quadratic Formula

Subtract 32 from both sides.

$$r^{2}-32=-4r$$

Add 4r to both sides.

$$r^{2}-32+4r=0$$

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.

$$r^{2}+4r-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$r=\frac{-4±\sqrt{16-4\left(-32\right)}}{2}$$

Multiply -4 times -32.

$$r=\frac{-4±\sqrt{16+128}}{2}$$

Add 16 to 128.

$$r=\frac{-4±\sqrt{144}}{2}$$

Take the square root of 144.

$$r=\frac{-4±12}{2}$$

Now solve the equation r=\frac{-4±12}{2} when ± is plus. Add -4 to 12.

$$r=\frac{8}{2}$$

Divide 8 by 2.

$$r=4$$

Now solve the equation r=\frac{-4±12}{2} when ± is minus. Subtract 12 from -4.

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

Divide -16 by 2.

$$r=-8$$

The equation is now solved.

$$r=4 r=-8$$

Steps for Completing the Square

Add 4r to both sides.

$$r^{2}+4r=32$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+4r+2^{2}=32+2^{2}$$

Square 2.

$$r^{2}+4r+4=32+4$$

Add 32 to 4.

$$r^{2}+4r+4=36$$

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

$$\left(r+2\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+2\right)^{2}}=\sqrt{36}$$

Simplify.

$$r+2=6 r+2=-6$$

Subtract 2 from both sides of the equation.

$$r=4 r=-8$$