Solve 3y^2+8y=12y+15

$$3 y ^ { 2 } + 8 y = 12 y + 15$$

Solve for y

$$y = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$y=3$$

Steps Using Factoring By Grouping

Subtract 12y from both sides.

$$3y^{2}+8y-12y=15$$

Combine 8y and -12y to get -4y.

$$3y^{2}-4y=15$$

Subtract 15 from both sides.

$$3y^{2}-4y-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3y^{2}+ay+by-15. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\left(-15\right)=-45$$

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 -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -4.

$$a=-9 b=5$$

Rewrite 3y^{2}-4y-15 as \left(3y^{2}-9y\right)+\left(5y-15\right).

$$\left(3y^{2}-9y\right)+\left(5y-15\right)$$

Factor out 3y in the first and 5 in the second group.

$$3y\left(y-3\right)+5\left(y-3\right)$$

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

$$\left(y-3\right)\left(3y+5\right)$$

To find equation solutions, solve y-3=0 and 3y+5=0.

$$y=3 y=-\frac{5}{3}$$

Steps Using the Quadratic Formula

Subtract 12y from both sides.

$$3y^{2}+8y-12y=15$$

Combine 8y and -12y to get -4y.

$$3y^{2}-4y=15$$

Subtract 15 from both sides.

$$3y^{2}-4y-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-4\right)±\sqrt{16-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$y=\frac{-\left(-4\right)±\sqrt{16+180}}{2\times 3}$$

Add 16 to 180.

$$y=\frac{-\left(-4\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$y=\frac{-\left(-4\right)±14}{2\times 3}$$

The opposite of -4 is 4.

$$y=\frac{4±14}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{4±14}{6}$$

Now solve the equation y=\frac{4±14}{6} when ± is plus. Add 4 to 14.

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

Divide 18 by 6.

$$y=3$$

Now solve the equation y=\frac{4±14}{6} when ± is minus. Subtract 14 from 4.

$$y=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$y=-\frac{5}{3}$$

The equation is now solved.

$$y=3 y=-\frac{5}{3}$$

Steps for Completing the Square

Subtract 12y from both sides.

$$3y^{2}+8y-12y=15$$

Combine 8y and -12y to get -4y.

$$3y^{2}-4y=15$$

Divide both sides by 3.

$$\frac{3y^{2}-4y}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{4}{3}y=\frac{15}{3}$$

Divide 15 by 3.

$$y^{2}-\frac{4}{3}y=5$$

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

$$y^{2}-\frac{4}{3}y+\left(-\frac{2}{3}\right)^{2}=5+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{4}{3}y+\frac{4}{9}=5+\frac{4}{9}$$

Add 5 to \frac{4}{9}.

$$y^{2}-\frac{4}{3}y+\frac{4}{9}=\frac{49}{9}$$

Factor y^{2}-\frac{4}{3}y+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{2}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{2}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$y-\frac{2}{3}=\frac{7}{3} y-\frac{2}{3}=-\frac{7}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$y=3 y=-\frac{5}{3}$$