Solve left(x-3right)^2=16

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

Solve for x

$$x=7$$

$$x=-1$$

Steps Using Factoring

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

$$x^{2}-6x+9=16$$

Subtract 16 from both sides.

$$x^{2}-6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}-6x-7=0$$

To solve the equation, factor x^{2}-6x-7 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.

$$a+b=-6 ab=-7$$

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. The only such pair is the system solution.

$$a=-7 b=1$$

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

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

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

$$x=7 x=-1$$

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}.

$$x^{2}-6x+9=16$$

Subtract 16 from both sides.

$$x^{2}-6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}-6x-7=0$$

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

$$a+b=-6 ab=1\left(-7\right)=-7$$

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. The only such pair is the system solution.

$$a=-7 b=1$$

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

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

Factor out x in x^{2}-7x.

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

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

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

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

$$x=7 x=-1$$

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}.

$$x^{2}-6x+9=16$$

Subtract 16 from both sides.

$$x^{2}-6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}-6x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-7\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

$$x=\frac{-\left(-6\right)±\sqrt{36+28}}{2}$$

Add 36 to 28.

$$x=\frac{-\left(-6\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-\left(-6\right)±8}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±8}{2}$$

Now solve the equation x=\frac{6±8}{2} when ± is plus. Add 6 to 8.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{6±8}{2} when ± is minus. Subtract 8 from 6.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=7 x=-1$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-3=4 x-3=-4$$

Add 3 to both sides of the equation.

$$x=7 x=-1$$