Solve 5r^2-30=5r

$$5 r ^ { 2 } - 30 = 5 r$$

Solve for r

$$r=-2$$

$$r=3$$

Steps Using Factoring By Grouping

Subtract 5r from both sides.

$$5r^{2}-30-5r=0$$

Divide both sides by 5.

$$r^{2}-6-r=0$$

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

$$r^{2}-r-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-6\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite r^{2}-r-6 as \left(r^{2}-3r\right)+\left(2r-6\right).

$$\left(r^{2}-3r\right)+\left(2r-6\right)$$

Factor out r in the first and 2 in the second group.

$$r\left(r-3\right)+2\left(r-3\right)$$

Factor out common term r-3 by using distributive property.

$$\left(r-3\right)\left(r+2\right)$$

To find equation solutions, solve r-3=0 and r+2=0.

$$r=3 r=-2$$

Steps Using the Quadratic Formula

Subtract 5r from both sides.

$$5r^{2}-30-5r=0$$

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.

$$5r^{2}-5r-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -5 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-30\right)}}{2\times 5}$$

Square -5.

$$r=\frac{-\left(-5\right)±\sqrt{25-4\times 5\left(-30\right)}}{2\times 5}$$

Multiply -4 times 5.

$$r=\frac{-\left(-5\right)±\sqrt{25-20\left(-30\right)}}{2\times 5}$$

Multiply -20 times -30.

$$r=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 5}$$

Add 25 to 600.

$$r=\frac{-\left(-5\right)±\sqrt{625}}{2\times 5}$$

Take the square root of 625.

$$r=\frac{-\left(-5\right)±25}{2\times 5}$$

The opposite of -5 is 5.

$$r=\frac{5±25}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{5±25}{10}$$

Now solve the equation r=\frac{5±25}{10} when ± is plus. Add 5 to 25.

$$r=\frac{30}{10}$$

Divide 30 by 10.

$$r=3$$

Now solve the equation r=\frac{5±25}{10} when ± is minus. Subtract 25 from 5.

$$r=-\frac{20}{10}$$

Divide -20 by 10.

$$r=-2$$

The equation is now solved.

$$r=3 r=-2$$

Steps for Completing the Square

Subtract 5r from both sides.

$$5r^{2}-30-5r=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$5r^{2}-5r=30$$

Divide both sides by 5.

$$\frac{5r^{2}-5r}{5}=\frac{30}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$r^{2}+\left(-\frac{5}{5}\right)r=\frac{30}{5}$$

Divide -5 by 5.

$$r^{2}-r=\frac{30}{5}$$

Divide 30 by 5.

$$r^{2}-r=6$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-r+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-r+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$r^{2}-r+\frac{1}{4}=\frac{25}{4}$$

Factor r^{2}-r+\frac{1}{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}.

$$\left(r-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$r=3 r=-2$$