Solve (x-5)^2-64=0

$$( x - 5 ) ^ { 2 } - 64 = 0$$

Solve for x

$$x=13$$

$$x=-3$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-64=0$$

Subtract 64 from 25 to get -39.

$$x^{2}-10x-39=0$$

To solve the equation, factor x^{2}-10x-39 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=-39$$

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 -39.

$$1,-39 3,-13$$

Calculate the sum for each pair.

$$1-39=-38 3-13=-10$$

The solution is the pair that gives sum -10.

$$a=-13 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-13\right)\left(x+3\right)$$

To find equation solutions, solve x-13=0 and x+3=0.

$$x=13 x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-64=0$$

Subtract 64 from 25 to get -39.

$$x^{2}-10x-39=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-39. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-39\right)=-39$$

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 -39.

$$1,-39 3,-13$$

Calculate the sum for each pair.

$$1-39=-38 3-13=-10$$

The solution is the pair that gives sum -10.

$$a=-13 b=3$$

Rewrite x^{2}-10x-39 as \left(x^{2}-13x\right)+\left(3x-39\right).

$$\left(x^{2}-13x\right)+\left(3x-39\right)$$

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

$$x\left(x-13\right)+3\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(x+3\right)$$

To find equation solutions, solve x-13=0 and x+3=0.

$$x=13 x=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-64=0$$

Subtract 64 from 25 to get -39.

$$x^{2}-10x-39=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-39\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-39\right)}}{2}$$

Multiply -4 times -39.

$$x=\frac{-\left(-10\right)±\sqrt{100+156}}{2}$$

Add 100 to 156.

$$x=\frac{-\left(-10\right)±\sqrt{256}}{2}$$

Take the square root of 256.

$$x=\frac{-\left(-10\right)±16}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±16}{2}$$

Now solve the equation x=\frac{10±16}{2} when ± is plus. Add 10 to 16.

$$x=\frac{26}{2}$$

Divide 26 by 2.

$$x=13$$

Now solve the equation x=\frac{10±16}{2} when ± is minus. Subtract 16 from 10.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=13 x=-3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25-64=0$$

Subtract 64 from 25 to get -39.

$$x^{2}-10x-39=0$$

Add 39 to both sides. Anything plus zero gives itself.

$$x^{2}-10x=39$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=39+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=39+25$$

Add 39 to 25.

$$x^{2}-10x+25=64$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-5=8 x-5=-8$$

Add 5 to both sides of the equation.

$$x=13 x=-3$$