Solve (2x-3)^2=16

$$( 2 x - 3 ) ^ { 2 } = 16$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

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

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}.

$$4x^{2}-12x+9=16$$

Subtract 16 from both sides.

$$4x^{2}-12x+9-16=0$$

Subtract 16 from 9 to get -7.

$$4x^{2}-12x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-12 ab=4\left(-7\right)=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -12.

$$a=-14 b=2$$

Rewrite 4x^{2}-12x-7 as \left(4x^{2}-14x\right)+\left(2x-7\right).

$$\left(4x^{2}-14x\right)+\left(2x-7\right)$$

Factor out 2x in 4x^{2}-14x.

$$2x\left(2x-7\right)+2x-7$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(2x+1\right)$$

To find equation solutions, solve 2x-7=0 and 2x+1=0.

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

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}.

$$4x^{2}-12x+9=16$$

Subtract 16 from both sides.

$$4x^{2}-12x+9-16=0$$

Subtract 16 from 9 to get -7.

$$4x^{2}-12x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-7\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-7\right)}}{2\times 4}$$

Multiply -16 times -7.

$$x=\frac{-\left(-12\right)±\sqrt{144+112}}{2\times 4}$$

Add 144 to 112.

$$x=\frac{-\left(-12\right)±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

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

The opposite of -12 is 12.

$$x=\frac{12±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±16}{8}$$

Now solve the equation x=\frac{12±16}{8} when ± is plus. Add 12 to 16.

$$x=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{12±16}{8} when ± is minus. Subtract 16 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.

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

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}.

$$4x^{2}-12x+9=16$$

Subtract 9 from both sides.

$$4x^{2}-12x=16-9$$

Subtract 9 from 16 to get 7.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{7}{4}$$

Divide -12 by 4.

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

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.

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

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

$$x^{2}-3x+\frac{9}{4}=\frac{7+9}{4}$$

Add \frac{7}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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