Solve 18+(3-x)^{2}+(x^{2}-2x-3)^2=x^2+(x^2-2x-3+3)^2

$$18 + ( 3 - x ) ^ { 2 } + ( x ^ { 2 } - 2 x - 3 ) ^ { 2 } = x ^ { 2 } + ( x ^ { 2 } - 2 x - 3 + 3 ) ^ { 2 }$$

Solve for x

$$x=-2$$

$$x=3$$

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}.

$$18+9-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 18 and 9 to get 27.

$$27-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Square x^{2}-2x-3.

$$27-6x+x^{2}+x^{4}-4x^{3}-2x^{2}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

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

$$27-6x-x^{2}+x^{4}-4x^{3}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Combine -6x and 12x to get 6x.

$$27+6x-x^{2}+x^{4}-4x^{3}+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 27 and 9 to get 36.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add -3 and 3 to get 0.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x\right)^{2}$$

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

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}\right)^{2}-4x^{2}x+4x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{2}x+4x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{3}+4x^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$36+6x-x^{2}+x^{4}-4x^{3}=5x^{2}+x^{4}-4x^{3}$$

Subtract 5x^{2} from both sides.

$$36+6x-x^{2}+x^{4}-4x^{3}-5x^{2}=x^{4}-4x^{3}$$

Combine -x^{2} and -5x^{2} to get -6x^{2}.

$$36+6x-6x^{2}+x^{4}-4x^{3}=x^{4}-4x^{3}$$

Subtract x^{4} from both sides.

$$36+6x-6x^{2}+x^{4}-4x^{3}-x^{4}=-4x^{3}$$

Combine x^{4} and -x^{4} to get 0.

$$36+6x-6x^{2}-4x^{3}=-4x^{3}$$

Add 4x^{3} to both sides.

$$36+6x-6x^{2}-4x^{3}+4x^{3}=0$$

Combine -4x^{3} and 4x^{3} to get 0.

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

Divide both sides by 6.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -x^{2}+x+6 as \left(-x^{2}+3x\right)+\left(-2x+6\right).

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

Factor out -x in the first and -2 in the second group.

$$-x\left(x-3\right)-2\left(x-3\right)$$

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

$$\left(x-3\right)\left(-x-2\right)$$

To find equation solutions, solve x-3=0 and -x-2=0.

$$x=3 x=-2$$

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}.

$$18+9-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 18 and 9 to get 27.

$$27-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Square x^{2}-2x-3.

$$27-6x+x^{2}+x^{4}-4x^{3}-2x^{2}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

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

$$27-6x-x^{2}+x^{4}-4x^{3}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Combine -6x and 12x to get 6x.

$$27+6x-x^{2}+x^{4}-4x^{3}+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 27 and 9 to get 36.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add -3 and 3 to get 0.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x\right)^{2}$$

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

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}\right)^{2}-4x^{2}x+4x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{2}x+4x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{3}+4x^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$36+6x-x^{2}+x^{4}-4x^{3}=5x^{2}+x^{4}-4x^{3}$$

Subtract 5x^{2} from both sides.

$$36+6x-x^{2}+x^{4}-4x^{3}-5x^{2}=x^{4}-4x^{3}$$

Combine -x^{2} and -5x^{2} to get -6x^{2}.

$$36+6x-6x^{2}+x^{4}-4x^{3}=x^{4}-4x^{3}$$

Subtract x^{4} from both sides.

$$36+6x-6x^{2}+x^{4}-4x^{3}-x^{4}=-4x^{3}$$

Combine x^{4} and -x^{4} to get 0.

$$36+6x-6x^{2}-4x^{3}=-4x^{3}$$

Add 4x^{3} to both sides.

$$36+6x-6x^{2}-4x^{3}+4x^{3}=0$$

Combine -4x^{3} and 4x^{3} to get 0.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 6 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 6.

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

Multiply -4 times -6.

$$x=\frac{-6±\sqrt{36+24\times 36}}{2\left(-6\right)}$$

Multiply 24 times 36.

$$x=\frac{-6±\sqrt{36+864}}{2\left(-6\right)}$$

Add 36 to 864.

$$x=\frac{-6±\sqrt{900}}{2\left(-6\right)}$$

Take the square root of 900.

$$x=\frac{-6±30}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-6±30}{-12}$$

Now solve the equation x=\frac{-6±30}{-12} when ± is plus. Add -6 to 30.

$$x=\frac{24}{-12}$$

Divide 24 by -12.

$$x=-2$$

Now solve the equation x=\frac{-6±30}{-12} when ± is minus. Subtract 30 from -6.

$$x=-\frac{36}{-12}$$

Divide -36 by -12.

$$x=3$$

The equation is now solved.

$$x=-2 x=3$$

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}.

$$18+9-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 18 and 9 to get 27.

$$27-6x+x^{2}+\left(x^{2}-2x-3\right)^{2}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Square x^{2}-2x-3.

$$27-6x+x^{2}+x^{4}-4x^{3}-2x^{2}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

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

$$27-6x-x^{2}+x^{4}-4x^{3}+12x+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Combine -6x and 12x to get 6x.

$$27+6x-x^{2}+x^{4}-4x^{3}+9=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add 27 and 9 to get 36.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x-3+3\right)^{2}$$

Add -3 and 3 to get 0.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}-2x\right)^{2}$$

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

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+\left(x^{2}\right)^{2}-4x^{2}x+4x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{2}x+4x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$36+6x-x^{2}+x^{4}-4x^{3}=x^{2}+x^{4}-4x^{3}+4x^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$36+6x-x^{2}+x^{4}-4x^{3}=5x^{2}+x^{4}-4x^{3}$$

Subtract 5x^{2} from both sides.

$$36+6x-x^{2}+x^{4}-4x^{3}-5x^{2}=x^{4}-4x^{3}$$

Combine -x^{2} and -5x^{2} to get -6x^{2}.

$$36+6x-6x^{2}+x^{4}-4x^{3}=x^{4}-4x^{3}$$

Subtract x^{4} from both sides.

$$36+6x-6x^{2}+x^{4}-4x^{3}-x^{4}=-4x^{3}$$

Combine x^{4} and -x^{4} to get 0.

$$36+6x-6x^{2}-4x^{3}=-4x^{3}$$

Add 4x^{3} to both sides.

$$36+6x-6x^{2}-4x^{3}+4x^{3}=0$$

Combine -4x^{3} and 4x^{3} to get 0.

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

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -6.

$$\frac{-6x^{2}+6x}{-6}=-\frac{36}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{6}{-6}x=-\frac{36}{-6}$$

Divide 6 by -6.

$$x^{2}-x=-\frac{36}{-6}$$

Divide -36 by -6.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$