Solve 3(2y-1)^2-(2y-1)=0

$$3 ( 2 y - 1 ) ^ { 2 } - ( 2 y - 1 ) = 0$$

Solve for y

$$y=\frac{2}{3}\approx 0.666666667$$

$$y=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-1\right)^{2}.

$$3\left(4y^{2}-4y+1\right)-\left(2y-1\right)=0$$

Use the distributive property to multiply 3 by 4y^{2}-4y+1.

$$12y^{2}-12y+3-\left(2y-1\right)=0$$

To find the opposite of 2y-1, find the opposite of each term.

$$12y^{2}-12y+3-2y+1=0$$

Combine -12y and -2y to get -14y.

$$12y^{2}-14y+3+1=0$$

Add 3 and 1 to get 4.

$$12y^{2}-14y+4=0$$

Divide both sides by 2.

$$6y^{2}-7y+2=0$$

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

$$a+b=-7 ab=6\times 2=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 6y^{2}-7y+2 as \left(6y^{2}-4y\right)+\left(-3y+2\right).

$$\left(6y^{2}-4y\right)+\left(-3y+2\right)$$

Factor out 2y in the first and -1 in the second group.

$$2y\left(3y-2\right)-\left(3y-2\right)$$

Factor out common term 3y-2 by using distributive property.

$$\left(3y-2\right)\left(2y-1\right)$$

To find equation solutions, solve 3y-2=0 and 2y-1=0.

$$y=\frac{2}{3} y=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-1\right)^{2}.

$$3\left(4y^{2}-4y+1\right)-\left(2y-1\right)=0$$

Use the distributive property to multiply 3 by 4y^{2}-4y+1.

$$12y^{2}-12y+3-\left(2y-1\right)=0$$

To find the opposite of 2y-1, find the opposite of each term.

$$12y^{2}-12y+3-2y+1=0$$

Combine -12y and -2y to get -14y.

$$12y^{2}-14y+3+1=0$$

Add 3 and 1 to get 4.

$$12y^{2}-14y+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -14 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 12\times 4}}{2\times 12}$$

Square -14.

$$y=\frac{-\left(-14\right)±\sqrt{196-4\times 12\times 4}}{2\times 12}$$

Multiply -4 times 12.

$$y=\frac{-\left(-14\right)±\sqrt{196-48\times 4}}{2\times 12}$$

Multiply -48 times 4.

$$y=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 12}$$

Add 196 to -192.

$$y=\frac{-\left(-14\right)±\sqrt{4}}{2\times 12}$$

Take the square root of 4.

$$y=\frac{-\left(-14\right)±2}{2\times 12}$$

The opposite of -14 is 14.

$$y=\frac{14±2}{2\times 12}$$

Multiply 2 times 12.

$$y=\frac{14±2}{24}$$

Now solve the equation y=\frac{14±2}{24} when ± is plus. Add 14 to 2.

$$y=\frac{16}{24}$$

Reduce the fraction \frac{16}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation y=\frac{14±2}{24} when ± is minus. Subtract 2 from 14.

$$y=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

$$y=\frac{1}{2}$$

The equation is now solved.

$$y=\frac{2}{3} y=\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-1\right)^{2}.

$$3\left(4y^{2}-4y+1\right)-\left(2y-1\right)=0$$

Use the distributive property to multiply 3 by 4y^{2}-4y+1.

$$12y^{2}-12y+3-\left(2y-1\right)=0$$

To find the opposite of 2y-1, find the opposite of each term.

$$12y^{2}-12y+3-2y+1=0$$

Combine -12y and -2y to get -14y.

$$12y^{2}-14y+3+1=0$$

Add 3 and 1 to get 4.

$$12y^{2}-14y+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$12y^{2}-14y=-4$$

Divide both sides by 12.

$$\frac{12y^{2}-14y}{12}=-\frac{4}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$y^{2}+\left(-\frac{14}{12}\right)y=-\frac{4}{12}$$

Reduce the fraction \frac{-14}{12} to lowest terms by extracting and canceling out 2.

$$y^{2}-\frac{7}{6}y=-\frac{4}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

$$y^{2}-\frac{7}{6}y=-\frac{1}{3}$$

Divide -\frac{7}{6}, the coefficient of the x term, by 2 to get -\frac{7}{12}. Then add the square of -\frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{7}{6}y+\left(-\frac{7}{12}\right)^{2}=-\frac{1}{3}+\left(-\frac{7}{12}\right)^{2}$$

Square -\frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{7}{6}y+\frac{49}{144}=-\frac{1}{3}+\frac{49}{144}$$

Add -\frac{1}{3} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{7}{6}y+\frac{49}{144}=\frac{1}{144}$$

Factor y^{2}-\frac{7}{6}y+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{7}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$y-\frac{7}{12}=\frac{1}{12} y-\frac{7}{12}=-\frac{1}{12}$$

Add \frac{7}{12} to both sides of the equation.

$$y=\frac{2}{3} y=\frac{1}{2}$$