Solve 6(13.5)={left(x-2*1/2right)}^2

$$6(13.5)= { \left(x-2 \times \frac{ 1 }{ 2 } \right) }^{ 2 }$$

Solve for x

$$x=10$$

$$x=-8$$

Steps Using Factoring

Multiply 6 and 13.5 to get 81.

$$81=\left(x-2\times \frac{1}{2}\right)^{2}$$

Multiply 2 and \frac{1}{2} to get 1.

$$81=\left(x-1\right)^{2}$$

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

$$81=x^{2}-2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-2x+1=81$$

Subtract 81 from both sides.

$$x^{2}-2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}-2x-80=0$$

To solve the equation, factor x^{2}-2x-80 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.

$$a+b=-2 ab=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -2.

$$a=-10 b=8$$

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

$$\left(x-10\right)\left(x+8\right)$$

To find equation solutions, solve x-10=0 and x+8=0.

$$x=10 x=-8$$

Steps Using Factoring By Grouping

Multiply 6 and 13.5 to get 81.

$$81=\left(x-2\times \frac{1}{2}\right)^{2}$$

Multiply 2 and \frac{1}{2} to get 1.

$$81=\left(x-1\right)^{2}$$

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

$$81=x^{2}-2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-2x+1=81$$

Subtract 81 from both sides.

$$x^{2}-2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}-2x-80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-80. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-80\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -2.

$$a=-10 b=8$$

Rewrite x^{2}-2x-80 as \left(x^{2}-10x\right)+\left(8x-80\right).

$$\left(x^{2}-10x\right)+\left(8x-80\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-10\right)+8\left(x-10\right)$$

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

$$\left(x-10\right)\left(x+8\right)$$

To find equation solutions, solve x-10=0 and x+8=0.

$$x=10 x=-8$$

Steps Using the Quadratic Formula

Multiply 6 and 13.5 to get 81.

$$81=\left(x-2\times \frac{1}{2}\right)^{2}$$

Multiply 2 and \frac{1}{2} to get 1.

$$81=\left(x-1\right)^{2}$$

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

$$81=x^{2}-2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-2x+1=81$$

Subtract 81 from both sides.

$$x^{2}-2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}-2x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-80\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-80\right)}}{2}$$

Multiply -4 times -80.

$$x=\frac{-\left(-2\right)±\sqrt{4+320}}{2}$$

Add 4 to 320.

$$x=\frac{-\left(-2\right)±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-\left(-2\right)±18}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±18}{2}$$

Now solve the equation x=\frac{2±18}{2} when ± is plus. Add 2 to 18.

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

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{2±18}{2} when ± is minus. Subtract 18 from 2.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=10 x=-8$$

Steps for Completing the Square

Multiply 6 and 13.5 to get 81.

$$81=\left(x-2\times \frac{1}{2}\right)^{2}$$

Multiply 2 and \frac{1}{2} to get 1.

$$81=\left(x-1\right)^{2}$$

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

$$81=x^{2}-2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-2x+1=81$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{81}$$

Simplify.

$$x-1=9 x-1=-9$$

Add 1 to both sides of the equation.

$$x=10 x=-8$$