Solve 2left(3-xright)^2-162=0

$$2 { \left(3-x \right) }^{ 2 } -162=0$$

Solve for x

$$x=-6$$

$$x=12$$

Steps Using Factoring By Grouping

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

$$2\left(9-6x+x^{2}\right)-162=0$$

Use the distributive property to multiply 2 by 9-6x+x^{2}.

$$18-12x+2x^{2}-162=0$$

Subtract 162 from 18 to get -144.

$$-144-12x+2x^{2}=0$$

Divide both sides by 2.

$$-72-6x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-6x-72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-72. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-72\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -6.

$$a=-12 b=6$$

Rewrite x^{2}-6x-72 as \left(x^{2}-12x\right)+\left(6x-72\right).

$$\left(x^{2}-12x\right)+\left(6x-72\right)$$

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

$$x\left(x-12\right)+6\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+6\right)$$

To find equation solutions, solve x-12=0 and x+6=0.

$$x=12 x=-6$$

Steps Using the Quadratic Formula

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

$$2\left(9-6x+x^{2}\right)-162=0$$

Use the distributive property to multiply 2 by 9-6x+x^{2}.

$$18-12x+2x^{2}-162=0$$

Subtract 162 from 18 to get -144.

$$-144-12x+2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-12x-144=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and -144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\left(-144\right)}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\left(-144\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\left(-144\right)}}{2\times 2}$$

Multiply -8 times -144.

$$x=\frac{-\left(-12\right)±\sqrt{144+1152}}{2\times 2}$$

Add 144 to 1152.

$$x=\frac{-\left(-12\right)±\sqrt{1296}}{2\times 2}$$

Take the square root of 1296.

$$x=\frac{-\left(-12\right)±36}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±36}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±36}{4}$$

Now solve the equation x=\frac{12±36}{4} when ± is plus. Add 12 to 36.

$$x=\frac{48}{4}$$

Divide 48 by 4.

$$x=12$$

Now solve the equation x=\frac{12±36}{4} when ± is minus. Subtract 36 from 12.

$$x=-\frac{24}{4}$$

Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=12 x=-6$$

Steps for Completing the Square

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

$$2\left(9-6x+x^{2}\right)-162=0$$

Use the distributive property to multiply 2 by 9-6x+x^{2}.

$$18-12x+2x^{2}-162=0$$

Subtract 162 from 18 to get -144.

$$-144-12x+2x^{2}=0$$

Add 144 to both sides. Anything plus zero gives itself.

$$-12x+2x^{2}=144$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-12x=144$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=\frac{144}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=\frac{144}{2}$$

Divide -12 by 2.

$$x^{2}-6x=\frac{144}{2}$$

Divide 144 by 2.

$$x^{2}-6x=72$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=72+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=72+9$$

Add 72 to 9.

$$x^{2}-6x+9=81$$

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}.

$$\left(x-3\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{81}$$

Simplify.

$$x-3=9 x-3=-9$$

Add 3 to both sides of the equation.

$$x=12 x=-6$$