Solve (x-3)^2-13=36

$$( x - 3 ) ^ { 2 } - 13 = 36$$

Solve for x

$$x=10$$

$$x=-4$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-13=36$$

Subtract 13 from 9 to get -4.

$$x^{2}-6x-4=36$$

Subtract 36 from both sides.

$$x^{2}-6x-4-36=0$$

Subtract 36 from -4 to get -40.

$$x^{2}-6x-40=0$$

To solve the equation, factor x^{2}-6x-40 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.

$$a+b=-6 ab=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -6.

$$a=-10 b=4$$

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

$$\left(x-10\right)\left(x+4\right)$$

To find equation solutions, solve x-10=0 and x+4=0.

$$x=10 x=-4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-13=36$$

Subtract 13 from 9 to get -4.

$$x^{2}-6x-4=36$$

Subtract 36 from both sides.

$$x^{2}-6x-4-36=0$$

Subtract 36 from -4 to get -40.

$$x^{2}-6x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-40\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -6.

$$a=-10 b=4$$

Rewrite x^{2}-6x-40 as \left(x^{2}-10x\right)+\left(4x-40\right).

$$\left(x^{2}-10x\right)+\left(4x-40\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-10\right)+4\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+4\right)$$

To find equation solutions, solve x-10=0 and x+4=0.

$$x=10 x=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-13=36$$

Subtract 13 from 9 to get -4.

$$x^{2}-6x-4=36$$

Subtract 36 from both sides.

$$x^{2}-6x-4-36=0$$

Subtract 36 from -4 to get -40.

$$x^{2}-6x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-40\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$x=\frac{-\left(-6\right)±\sqrt{36+160}}{2}$$

Add 36 to 160.

$$x=\frac{-\left(-6\right)±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-\left(-6\right)±14}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±14}{2}$$

Now solve the equation x=\frac{6±14}{2} when ± is plus. Add 6 to 14.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{6±14}{2} when ± is minus. Subtract 14 from 6.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=10 x=-4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-13=36$$

Subtract 13 from 9 to get -4.

$$x^{2}-6x-4=36$$

Add 4 to both sides.

$$x^{2}-6x=36+4$$

Add 36 and 4 to get 40.

$$x^{2}-6x=40$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=40+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=40+9$$

Add 40 to 9.

$$x^{2}-6x+9=49$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{49}$$

Simplify.

$$x-3=7 x-3=-7$$

Add 3 to both sides of the equation.

$$x=10 x=-4$$