Solve (2x-3)^2-4=0

$$( 2 x - 3 ) ^ { 2 } - 4 = 0$$

Solve for x

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

$$x=\frac{1}{2}=0.5$$

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}.

$$4x^{2}-12x+9-4=0$$

Subtract 4 from 9 to get 5.

$$4x^{2}-12x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-12 ab=4\times 5=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-10 b=-2$$

Rewrite 4x^{2}-12x+5 as \left(4x^{2}-10x\right)+\left(-2x+5\right).

$$\left(4x^{2}-10x\right)+\left(-2x+5\right)$$

Factor out 2x in the first and -1 in the second group.

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

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

$$\left(2x-5\right)\left(2x-1\right)$$

To find equation solutions, solve 2x-5=0 and 2x-1=0.

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

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}.

$$4x^{2}-12x+9-4=0$$

Subtract 4 from 9 to get 5.

$$4x^{2}-12x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 5}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2\times 4}$$

Add 144 to -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-\left(-12\right)±8}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±8}{8}$$

Now solve the equation x=\frac{12±8}{8} when ± is plus. Add 12 to 8.

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

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{12±8}{8} when ± is minus. Subtract 8 from 12.

$$x=\frac{4}{8}$$

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

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

The equation is now solved.

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

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}.

$$4x^{2}-12x+9-4=0$$

Subtract 4 from 9 to get 5.

$$4x^{2}-12x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-12x=-5$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=-\frac{5}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{5}{4}$$

Divide -12 by 4.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{5}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{1}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

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