Solve 3y^2+8y=2y+9

$$3 y ^ { 2 } + 8 y = 2 y + 9$$

Solve for y

$$y=-3$$

$$y=1$$

Steps Using Factoring By Grouping

Subtract 2y from both sides.

$$3y^{2}+8y-2y=9$$

Combine 8y and -2y to get 6y.

$$3y^{2}+6y=9$$

Subtract 9 from both sides.

$$3y^{2}+6y-9=0$$

Divide both sides by 3.

$$y^{2}+2y-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

$$y\left(y-1\right)+3\left(y-1\right)$$

Factor out common term y-1 by using distributive property.

$$\left(y-1\right)\left(y+3\right)$$

To find equation solutions, solve y-1=0 and y+3=0.

$$y=1 y=-3$$

Steps Using the Quadratic Formula

Subtract 2y from both sides.

$$3y^{2}+8y-2y=9$$

Combine 8y and -2y to get 6y.

$$3y^{2}+6y=9$$

Subtract 9 from both sides.

$$3y^{2}+6y-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-6±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$y=\frac{-6±\sqrt{36+108}}{2\times 3}$$

Add 36 to 108.

$$y=\frac{-6±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$y=\frac{-6±12}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-6±12}{6}$$

Now solve the equation y=\frac{-6±12}{6} when ± is plus. Add -6 to 12.

$$y=\frac{6}{6}$$

Divide 6 by 6.

$$y=1$$

Now solve the equation y=\frac{-6±12}{6} when ± is minus. Subtract 12 from -6.

$$y=-\frac{18}{6}$$

Divide -18 by 6.

$$y=-3$$

The equation is now solved.

$$y=1 y=-3$$

Steps for Completing the Square

Subtract 2y from both sides.

$$3y^{2}+8y-2y=9$$

Combine 8y and -2y to get 6y.

$$3y^{2}+6y=9$$

Divide both sides by 3.

$$\frac{3y^{2}+6y}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\frac{6}{3}y=\frac{9}{3}$$

Divide 6 by 3.

$$y^{2}+2y=\frac{9}{3}$$

Divide 9 by 3.

$$y^{2}+2y=3$$

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

$$y^{2}+2y+1^{2}=3+1^{2}$$

Square 1.

$$y^{2}+2y+1=3+1$$

Add 3 to 1.

$$y^{2}+2y+1=4$$

Factor y^{2}+2y+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$y+1=2 y+1=-2$$

Subtract 1 from both sides of the equation.

$$y=1 y=-3$$