Solve 20=9+x^2+left(4-xright)^2+1

$$20=9+ { x }^{ 2 } + { \left(4-x \right) }^{ 2 } +1$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$20=9+x^{2}+16-8x+x^{2}+1$$

Add 9 and 16 to get 25.

$$20=25+x^{2}-8x+x^{2}+1$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$20=25+2x^{2}-8x+1$$

Add 25 and 1 to get 26.

$$20=26+2x^{2}-8x$$

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

$$26+2x^{2}-8x=20$$

Subtract 20 from both sides.

$$26+2x^{2}-8x-20=0$$

Subtract 20 from 26 to get 6.

$$6+2x^{2}-8x=0$$

Divide both sides by 2.

$$3+x^{2}-4x=0$$

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

$$x^{2}-4x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite x^{2}-4x+3 as \left(x^{2}-3x\right)+\left(-x+3\right).

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

Factor out x in the first and -1 in the second group.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-1\right)$$

To find equation solutions, solve x-3=0 and x-1=0.

$$x=3 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$20=9+x^{2}+16-8x+x^{2}+1$$

Add 9 and 16 to get 25.

$$20=25+x^{2}-8x+x^{2}+1$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$20=25+2x^{2}-8x+1$$

Add 25 and 1 to get 26.

$$20=26+2x^{2}-8x$$

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

$$26+2x^{2}-8x=20$$

Subtract 20 from both sides.

$$26+2x^{2}-8x-20=0$$

Subtract 20 from 26 to get 6.

$$6+2x^{2}-8x=0$$

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.

$$2x^{2}-8x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 2}$$

Add 64 to -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$x=\frac{-\left(-8\right)±4}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4}{4}$$

Now solve the equation x=\frac{8±4}{4} when ± is plus. Add 8 to 4.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{8±4}{4} when ± is minus. Subtract 4 from 8.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=3 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$20=9+x^{2}+16-8x+x^{2}+1$$

Add 9 and 16 to get 25.

$$20=25+x^{2}-8x+x^{2}+1$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$20=25+2x^{2}-8x+1$$

Add 25 and 1 to get 26.

$$20=26+2x^{2}-8x$$

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

$$26+2x^{2}-8x=20$$

Subtract 26 from both sides.

$$2x^{2}-8x=20-26$$

Subtract 26 from 20 to get -6.

$$2x^{2}-8x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=-\frac{6}{2}$$

Divide -8 by 2.

$$x^{2}-4x=-\frac{6}{2}$$

Divide -6 by 2.

$$x^{2}-4x=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$