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

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

Solve for x

$$x=1$$

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

Steps Using Factoring By Grouping

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

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

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

$$3-12x+12x^{2}+1-2x=2$$

Add 3 and 1 to get 4.

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

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

$$4-14x+12x^{2}=2$$

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

$$2-14x+12x^{2}=0$$

Divide both sides by 2.

$$1-7x+6x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$6x^{2}-7x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-7 ab=6\times 1=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 6x^{2}-7x+1 as \left(6x^{2}-6x\right)+\left(-x+1\right).

$$\left(6x^{2}-6x\right)+\left(-x+1\right)$$

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

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

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

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

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

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

Steps Using the Quadratic Formula

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

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

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

$$3-12x+12x^{2}+1-2x=2$$

Add 3 and 1 to get 4.

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

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

$$4-14x+12x^{2}=2$$

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

$$2-14x+12x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$12x^{2}-14x+2=0$$

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

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

Square -14.

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

Multiply -4 times 12.

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

Multiply -48 times 2.

$$x=\frac{-\left(-14\right)±\sqrt{196-96}}{2\times 12}$$

Add 196 to -96.

$$x=\frac{-\left(-14\right)±\sqrt{100}}{2\times 12}$$

Take the square root of 100.

$$x=\frac{-\left(-14\right)±10}{2\times 12}$$

The opposite of -14 is 14.

$$x=\frac{14±10}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{14±10}{24}$$

Now solve the equation x=\frac{14±10}{24} when ± is plus. Add 14 to 10.

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

Divide 24 by 24.

$$x=1$$

Now solve the equation x=\frac{14±10}{24} when ± is minus. Subtract 10 from 14.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

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

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

$$3-12x+12x^{2}+1-2x=2$$

Add 3 and 1 to get 4.

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

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

$$4-14x+12x^{2}=2$$

Subtract 4 from both sides.

$$-14x+12x^{2}=2-4$$

Subtract 4 from 2 to get -2.

$$-14x+12x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$12x^{2}-14x=-2$$

Divide both sides by 12.

$$\frac{12x^{2}-14x}{12}=-\frac{2}{12}$$

Dividing by 12 undoes the multiplication by 12.

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

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

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

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

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

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.

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

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

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

Add -\frac{1}{6} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{25}{144}$$

Factor x^{2}-\frac{7}{6}x+\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}.

$$\left(x-\frac{7}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$x-\frac{7}{12}=\frac{5}{12} x-\frac{7}{12}=-\frac{5}{12}$$

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

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