Solve left(2x-3right)^2=2(2x-3)

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

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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

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}.

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

Use the distributive property to multiply 2 by 2x-3.

$$4x^{2}-12x+9=4x-6$$

Subtract 4x from both sides.

$$4x^{2}-12x+9-4x=-6$$

Combine -12x and -4x to get -16x.

$$4x^{2}-16x+9=-6$$

Add 6 to both sides.

$$4x^{2}-16x+9+6=0$$

Add 9 and 6 to get 15.

$$4x^{2}-16x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-16 ab=4\times 15=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -16.

$$a=-10 b=-6$$

Rewrite 4x^{2}-16x+15 as \left(4x^{2}-10x\right)+\left(-6x+15\right).

$$\left(4x^{2}-10x\right)+\left(-6x+15\right)$$

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

$$2x\left(2x-5\right)-3\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x-3\right)$$

To find equation solutions, solve 2x-5=0 and 2x-3=0.

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

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}.

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

Use the distributive property to multiply 2 by 2x-3.

$$4x^{2}-12x+9=4x-6$$

Subtract 4x from both sides.

$$4x^{2}-12x+9-4x=-6$$

Combine -12x and -4x to get -16x.

$$4x^{2}-16x+9=-6$$

Add 6 to both sides.

$$4x^{2}-16x+9+6=0$$

Add 9 and 6 to get 15.

$$4x^{2}-16x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 15}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 4}$$

Add 256 to -240.

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

Take the square root of 16.

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

The opposite of -16 is 16.

$$x=\frac{16±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±4}{8}$$

Now solve the equation x=\frac{16±4}{8} when ± is plus. Add 16 to 4.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{16±4}{8} when ± is minus. Subtract 4 from 16.

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

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

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

The equation is now solved.

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

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}.

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

Use the distributive property to multiply 2 by 2x-3.

$$4x^{2}-12x+9=4x-6$$

Subtract 4x from both sides.

$$4x^{2}-12x+9-4x=-6$$

Combine -12x and -4x to get -16x.

$$4x^{2}-16x+9=-6$$

Subtract 9 from both sides.

$$4x^{2}-16x=-6-9$$

Subtract 9 from -6 to get -15.

$$4x^{2}-16x=-15$$

Divide both sides by 4.

$$\frac{4x^{2}-16x}{4}=-\frac{15}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{16}{4}\right)x=-\frac{15}{4}$$

Divide -16 by 4.

$$x^{2}-4x=-\frac{15}{4}$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{15}{4}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-\frac{15}{4}+4$$

Add -\frac{15}{4} to 4.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=\frac{1}{2} x-2=-\frac{1}{2}$$

Add 2 to both sides of the equation.

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