Solve 2r^2-27=3r

$$2 r ^ { 2 } - 27 = 3 r$$

Solve for r

$$r=-3$$

$$r = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Subtract 3r from both sides.

$$2r^{2}-27-3r=0$$

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

$$2r^{2}-3r-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2r^{2}+ar+br-27. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-27\right)=-54$$

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 -54.

$$1,-54 2,-27 3,-18 6,-9$$

Calculate the sum for each pair.

$$1-54=-53 2-27=-25 3-18=-15 6-9=-3$$

The solution is the pair that gives sum -3.

$$a=-9 b=6$$

Rewrite 2r^{2}-3r-27 as \left(2r^{2}-9r\right)+\left(6r-27\right).

$$\left(2r^{2}-9r\right)+\left(6r-27\right)$$

Factor out r in the first and 3 in the second group.

$$r\left(2r-9\right)+3\left(2r-9\right)$$

Factor out common term 2r-9 by using distributive property.

$$\left(2r-9\right)\left(r+3\right)$$

To find equation solutions, solve 2r-9=0 and r+3=0.

$$r=\frac{9}{2} r=-3$$

Steps Using the Quadratic Formula

Subtract 3r from both sides.

$$2r^{2}-27-3r=0$$

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.

$$2r^{2}-3r-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-27\right)}}{2\times 2}$$

Square -3.

$$r=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-27\right)}}{2\times 2}$$

Multiply -4 times 2.

$$r=\frac{-\left(-3\right)±\sqrt{9-8\left(-27\right)}}{2\times 2}$$

Multiply -8 times -27.

$$r=\frac{-\left(-3\right)±\sqrt{9+216}}{2\times 2}$$

Add 9 to 216.

$$r=\frac{-\left(-3\right)±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$r=\frac{-\left(-3\right)±15}{2\times 2}$$

The opposite of -3 is 3.

$$r=\frac{3±15}{2\times 2}$$

Multiply 2 times 2.

$$r=\frac{3±15}{4}$$

Now solve the equation r=\frac{3±15}{4} when ± is plus. Add 3 to 15.

$$r=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$r=\frac{9}{2}$$

Now solve the equation r=\frac{3±15}{4} when ± is minus. Subtract 15 from 3.

$$r=-\frac{12}{4}$$

Divide -12 by 4.

$$r=-3$$

The equation is now solved.

$$r=\frac{9}{2} r=-3$$

Steps for Completing the Square

Subtract 3r from both sides.

$$2r^{2}-27-3r=0$$

Add 27 to both sides. Anything plus zero gives itself.

$$2r^{2}-3r=27$$

Divide both sides by 2.

$$\frac{2r^{2}-3r}{2}=\frac{27}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$r^{2}-\frac{3}{2}r=\frac{27}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{3}{2}r+\left(-\frac{3}{4}\right)^{2}=\frac{27}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{3}{2}r+\frac{9}{16}=\frac{27}{2}+\frac{9}{16}$$

Add \frac{27}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{3}{2}r+\frac{9}{16}=\frac{225}{16}$$

Factor r^{2}-\frac{3}{2}r+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{3}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{3}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$r-\frac{3}{4}=\frac{15}{4} r-\frac{3}{4}=-\frac{15}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$r=\frac{9}{2} r=-3$$