Solve 3y(y-2)=4y-8

$$3 y ( y - 2 ) = 4 y - 8$$

Solve for y

$$y=2$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3y by y-2.

$$3y^{2}-6y=4y-8$$

Subtract 4y from both sides.

$$3y^{2}-6y-4y=-8$$

Combine -6y and -4y to get -10y.

$$3y^{2}-10y=-8$$

Add 8 to both sides.

$$3y^{2}-10y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-10\right)±\sqrt{100-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$y=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 3}$$

Add 100 to -96.

$$y=\frac{-\left(-10\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$y=\frac{-\left(-10\right)±2}{2\times 3}$$

The opposite of -10 is 10.

$$y=\frac{10±2}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{10±2}{6}$$

Now solve the equation y=\frac{10±2}{6} when ± is plus. Add 10 to 2.

$$y=\frac{12}{6}$$

Divide 12 by 6.

$$y=2$$

Now solve the equation y=\frac{10±2}{6} when ± is minus. Subtract 2 from 10.

$$y=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{4}{3}$$

The equation is now solved.

$$y=2 y=\frac{4}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3y by y-2.

$$3y^{2}-6y=4y-8$$

Subtract 4y from both sides.

$$3y^{2}-6y-4y=-8$$

Combine -6y and -4y to get -10y.

$$3y^{2}-10y=-8$$

Divide both sides by 3.

$$\frac{3y^{2}-10y}{3}=-\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{10}{3}y=-\frac{8}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{10}{3}y+\left(-\frac{5}{3}\right)^{2}=-\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=-\frac{8}{3}+\frac{25}{9}$$

Add -\frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=\frac{1}{9}$$

Factor y^{2}-\frac{10}{3}y+\frac{25}{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}.

$$\left(y-\frac{5}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$y-\frac{5}{3}=\frac{1}{3} y-\frac{5}{3}=-\frac{1}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$y=2 y=\frac{4}{3}$$