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

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

Solve for x

$$x=2$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=3x-1-\left(x-1\right)-3$$

To find the opposite of x-1, find the opposite of each term.

$$4x^{2}-12x+9=3x-1-x+1-3$$

Combine 3x and -x to get 2x.

$$4x^{2}-12x+9=2x-1+1-3$$

Add -1 and 1 to get 0.

$$4x^{2}-12x+9=2x-3$$

Subtract 2x from both sides.

$$4x^{2}-12x+9-2x=-3$$

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

$$4x^{2}-14x+9=-3$$

Add 3 to both sides.

$$4x^{2}-14x+9+3=0$$

Add 9 and 3 to get 12.

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

Divide both sides by 2.

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

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

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

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

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

Factor out 2x in the first and -3 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=3x-1-\left(x-1\right)-3$$

To find the opposite of x-1, find the opposite of each term.

$$4x^{2}-12x+9=3x-1-x+1-3$$

Combine 3x and -x to get 2x.

$$4x^{2}-12x+9=2x-1+1-3$$

Add -1 and 1 to get 0.

$$4x^{2}-12x+9=2x-3$$

Subtract 2x from both sides.

$$4x^{2}-12x+9-2x=-3$$

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

$$4x^{2}-14x+9=-3$$

Add 3 to both sides.

$$4x^{2}-14x+9+3=0$$

Add 9 and 3 to get 12.

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

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

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

Square -14.

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

Multiply -4 times 4.

$$x=\frac{-\left(-14\right)±\sqrt{196-16\times 12}}{2\times 4}$$

Multiply -16 times 12.

$$x=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 4}$$

Add 196 to -192.

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

Take the square root of 4.

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

The opposite of -14 is 14.

$$x=\frac{14±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{14±2}{8}$$

Now solve the equation x=\frac{14±2}{8} when ± is plus. Add 14 to 2.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{14±2}{8} when ± is minus. Subtract 2 from 14.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=3x-1-\left(x-1\right)-3$$

To find the opposite of x-1, find the opposite of each term.

$$4x^{2}-12x+9=3x-1-x+1-3$$

Combine 3x and -x to get 2x.

$$4x^{2}-12x+9=2x-1+1-3$$

Add -1 and 1 to get 0.

$$4x^{2}-12x+9=2x-3$$

Subtract 2x from both sides.

$$4x^{2}-12x+9-2x=-3$$

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

$$4x^{2}-14x+9=-3$$

Subtract 9 from both sides.

$$4x^{2}-14x=-3-9$$

Subtract 9 from -3 to get -12.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

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

Divide -12 by 4.

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

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

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

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-3+\frac{49}{16}$$

Add -3 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{1}{4} x-\frac{7}{4}=-\frac{1}{4}$$

Add \frac{7}{4} to both sides of the equation.

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