Solve (x-1)^2+(2-3x)(1-x)=0

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

Solve for x

$$x=1$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 2-3x by 1-x and combine like terms.

$$x^{2}-2x+1+2-5x+3x^{2}=0$$

Add 1 and 2 to get 3.

$$x^{2}-2x+3-5x+3x^{2}=0$$

Combine -2x and -5x to get -7x.

$$x^{2}-7x+3+3x^{2}=0$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-7x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=4\times 3=12$$

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

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

Factor out 4x in the first and -3 in the second group.

$$4x\left(x-1\right)-3\left(x-1\right)$$

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

$$\left(x-1\right)\left(4x-3\right)$$

To find equation solutions, solve x-1=0 and 4x-3=0.

$$x=1 x=\frac{3}{4}$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 2-3x by 1-x and combine like terms.

$$x^{2}-2x+1+2-5x+3x^{2}=0$$

Add 1 and 2 to get 3.

$$x^{2}-2x+3-5x+3x^{2}=0$$

Combine -2x and -5x to get -7x.

$$x^{2}-7x+3+3x^{2}=0$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-7x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-7\right)±\sqrt{49-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 4}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2\times 4}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2\times 4}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{7±1}{8}$$

Now solve the equation x=\frac{7±1}{8} when ± is plus. Add 7 to 1.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{7±1}{8} when ± is minus. Subtract 1 from 7.

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

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

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

The equation is now solved.

$$x=1 x=\frac{3}{4}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply 2-3x by 1-x and combine like terms.

$$x^{2}-2x+1+2-5x+3x^{2}=0$$

Add 1 and 2 to get 3.

$$x^{2}-2x+3-5x+3x^{2}=0$$

Combine -2x and -5x to get -7x.

$$x^{2}-7x+3+3x^{2}=0$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-7x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-7x=-3$$

Divide both sides by 4.

$$\frac{4x^{2}-7x}{4}=-\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{7}{4}x=-\frac{3}{4}$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=-\frac{3}{4}+\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=-\frac{3}{4}+\frac{49}{64}$$

Add -\frac{3}{4} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{7}{8}=\frac{1}{8} x-\frac{7}{8}=-\frac{1}{8}$$

Add \frac{7}{8} to both sides of the equation.

$$x=1 x=\frac{3}{4}$$