Solve (2x-1)^2-169=0

$$( 2 x - 1 ) ^ { 2 } - 169 = 0$$

Solve for x

$$x=7$$

$$x=-6$$

Steps Using Factoring By Grouping

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

$$4x^{2}-4x+1-169=0$$

Subtract 169 from 1 to get -168.

$$4x^{2}-4x-168=0$$

Divide both sides by 4.

$$x^{2}-x-42=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-42. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-42\right)=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=-7 b=6$$

Rewrite x^{2}-x-42 as \left(x^{2}-7x\right)+\left(6x-42\right).

$$\left(x^{2}-7x\right)+\left(6x-42\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-7\right)+6\left(x-7\right)$$

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

$$\left(x-7\right)\left(x+6\right)$$

To find equation solutions, solve x-7=0 and x+6=0.

$$x=7 x=-6$$

Steps Using the Quadratic Formula

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

$$4x^{2}-4x+1-169=0$$

Subtract 169 from 1 to get -168.

$$4x^{2}-4x-168=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -168 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-168\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-168\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-168\right)}}{2\times 4}$$

Multiply -16 times -168.

$$x=\frac{-\left(-4\right)±\sqrt{16+2688}}{2\times 4}$$

Add 16 to 2688.

$$x=\frac{-\left(-4\right)±\sqrt{2704}}{2\times 4}$$

Take the square root of 2704.

$$x=\frac{-\left(-4\right)±52}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±52}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±52}{8}$$

Now solve the equation x=\frac{4±52}{8} when ± is plus. Add 4 to 52.

$$x=\frac{56}{8}$$

Divide 56 by 8.

$$x=7$$

Now solve the equation x=\frac{4±52}{8} when ± is minus. Subtract 52 from 4.

$$x=-\frac{48}{8}$$

Divide -48 by 8.

$$x=-6$$

The equation is now solved.

$$x=7 x=-6$$

Steps for Completing the Square

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

$$4x^{2}-4x+1-169=0$$

Subtract 169 from 1 to get -168.

$$4x^{2}-4x-168=0$$

Add 168 to both sides. Anything plus zero gives itself.

$$4x^{2}-4x=168$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=\frac{168}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{168}{4}$$

Divide -4 by 4.

$$x^{2}-x=\frac{168}{4}$$

Divide 168 by 4.

$$x^{2}-x=42$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=42+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=42+\frac{1}{4}$$

Add 42 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{169}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{13}{2} x-\frac{1}{2}=-\frac{13}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=7 x=-6$$