Solve x^2+(x-3)^2=5+8(2-x)

$$x ^ { 2 } + ( x - 3 ) ^ { 2 } = 5 + 8 ( 2 - x )$$

Solve for x

$$x=-3$$

$$x=2$$

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}.

$$x^{2}+x^{2}-6x+9=5+8\left(2-x\right)$$

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

$$2x^{2}-6x+9=5+8\left(2-x\right)$$

Use the distributive property to multiply 8 by 2-x.

$$2x^{2}-6x+9=5+16-8x$$

Add 5 and 16 to get 21.

$$2x^{2}-6x+9=21-8x$$

Subtract 21 from both sides.

$$2x^{2}-6x+9-21=-8x$$

Subtract 21 from 9 to get -12.

$$2x^{2}-6x-12=-8x$$

Add 8x to both sides.

$$2x^{2}-6x-12+8x=0$$

Combine -6x and 8x to get 2x.

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

Divide both sides by 2.

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

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.

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

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.

$$a=-2 b=3$$

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

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

Factor out x in the first and 3 in the second group.

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

Factor out common term x-2 by using distributive property.

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

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

$$x=2 x=-3$$

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}.

$$x^{2}+x^{2}-6x+9=5+8\left(2-x\right)$$

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

$$2x^{2}-6x+9=5+8\left(2-x\right)$$

Use the distributive property to multiply 8 by 2-x.

$$2x^{2}-6x+9=5+16-8x$$

Add 5 and 16 to get 21.

$$2x^{2}-6x+9=21-8x$$

Subtract 21 from both sides.

$$2x^{2}-6x+9-21=-8x$$

Subtract 21 from 9 to get -12.

$$2x^{2}-6x-12=-8x$$

Add 8x to both sides.

$$2x^{2}-6x-12+8x=0$$

Combine -6x and 8x to get 2x.

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

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

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

Square 2.

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

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-2±\sqrt{4+96}}{2\times 2}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±10}{4}$$

Now solve the equation x=\frac{-2±10}{4} when ± is plus. Add -2 to 10.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-2±10}{4} when ± is minus. Subtract 10 from -2.

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

Divide -12 by 4.

$$x=-3$$

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(x-3\right)^{2}.

$$x^{2}+x^{2}-6x+9=5+8\left(2-x\right)$$

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

$$2x^{2}-6x+9=5+8\left(2-x\right)$$

Use the distributive property to multiply 8 by 2-x.

$$2x^{2}-6x+9=5+16-8x$$

Add 5 and 16 to get 21.

$$2x^{2}-6x+9=21-8x$$

Add 8x to both sides.

$$2x^{2}-6x+9+8x=21$$

Combine -6x and 8x to get 2x.

$$2x^{2}+2x+9=21$$

Subtract 9 from both sides.

$$2x^{2}+2x=21-9$$

Subtract 9 from 21 to get 12.

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

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{12}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{12}{2}$$

Divide 12 by 2.

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

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.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=2 x=-3$$