Solve 3(2x-1)^2-4(2x-1)=4

$$3 ( 2 x - 1 ) ^ { 2 } - 4 ( 2 x - 1 ) = 4$$

Solve for x

$$x=\frac{1}{6}\approx 0.166666667$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

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

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

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

$$12x^{2}-12x+3-4\left(2x-1\right)=4$$

Use the distributive property to multiply -4 by 2x-1.

$$12x^{2}-12x+3-8x+4=4$$

Combine -12x and -8x to get -20x.

$$12x^{2}-20x+3+4=4$$

Add 3 and 4 to get 7.

$$12x^{2}-20x+7=4$$

Subtract 4 from both sides.

$$12x^{2}-20x+7-4=0$$

Subtract 4 from 7 to get 3.

$$12x^{2}-20x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-20 ab=12\times 3=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -20.

$$a=-18 b=-2$$

Rewrite 12x^{2}-20x+3 as \left(12x^{2}-18x\right)+\left(-2x+3\right).

$$\left(12x^{2}-18x\right)+\left(-2x+3\right)$$

Factor out 6x in the first and -1 in the second group.

$$6x\left(2x-3\right)-\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(6x-1\right)$$

To find equation solutions, solve 2x-3=0 and 6x-1=0.

$$x=\frac{3}{2} x=\frac{1}{6}$$

Steps Using the Quadratic Formula

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

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

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

$$12x^{2}-12x+3-4\left(2x-1\right)=4$$

Use the distributive property to multiply -4 by 2x-1.

$$12x^{2}-12x+3-8x+4=4$$

Combine -12x and -8x to get -20x.

$$12x^{2}-20x+3+4=4$$

Add 3 and 4 to get 7.

$$12x^{2}-20x+7=4$$

Subtract 4 from both sides.

$$12x^{2}-20x+7-4=0$$

Subtract 4 from 7 to get 3.

$$12x^{2}-20x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -20 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 12\times 3}}{2\times 12}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 12\times 3}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-20\right)±\sqrt{400-48\times 3}}{2\times 12}$$

Multiply -48 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-144}}{2\times 12}$$

Add 400 to -144.

$$x=\frac{-\left(-20\right)±\sqrt{256}}{2\times 12}$$

Take the square root of 256.

$$x=\frac{-\left(-20\right)±16}{2\times 12}$$

The opposite of -20 is 20.

$$x=\frac{20±16}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{20±16}{24}$$

Now solve the equation x=\frac{20±16}{24} when ± is plus. Add 20 to 16.

$$x=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{20±16}{24} when ± is minus. Subtract 16 from 20.

$$x=\frac{4}{24}$$

Reduce the fraction \frac{4}{24} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{6}$$

The equation is now solved.

$$x=\frac{3}{2} x=\frac{1}{6}$$

Steps for Completing the Square

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

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

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

$$12x^{2}-12x+3-4\left(2x-1\right)=4$$

Use the distributive property to multiply -4 by 2x-1.

$$12x^{2}-12x+3-8x+4=4$$

Combine -12x and -8x to get -20x.

$$12x^{2}-20x+3+4=4$$

Add 3 and 4 to get 7.

$$12x^{2}-20x+7=4$$

Subtract 7 from both sides.

$$12x^{2}-20x=4-7$$

Subtract 7 from 4 to get -3.

$$12x^{2}-20x=-3$$

Divide both sides by 12.

$$\frac{12x^{2}-20x}{12}=-\frac{3}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{20}{12}\right)x=-\frac{3}{12}$$

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{5}{3}x=-\frac{3}{12}$$

Reduce the fraction \frac{-3}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{5}{3}x=-\frac{1}{4}$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{1}{4}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{1}{4}+\frac{25}{36}$$

Add -\frac{1}{4} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{4}{9}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x-\frac{5}{6}=\frac{2}{3} x-\frac{5}{6}=-\frac{2}{3}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{1}{6}$$