Solve x^2-6x+9=(5-2x)^2

$$x ^ { 2 } - 6 x + 9 = ( 5 - 2 x ) ^ { 2 }$$

Solve for x

$$x=2$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-2x\right)^{2}.

$$x^{2}-6x+9=25-20x+4x^{2}$$

Subtract 25 from both sides.

$$x^{2}-6x+9-25=-20x+4x^{2}$$

Subtract 25 from 9 to get -16.

$$x^{2}-6x-16=-20x+4x^{2}$$

Add 20x to both sides.

$$x^{2}-6x-16+20x=4x^{2}$$

Combine -6x and 20x to get 14x.

$$x^{2}+14x-16=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+14x-16-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+14x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=14 ab=-3\left(-16\right)=48$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 48.

$$1,48 2,24 3,16 4,12 6,8$$

Calculate the sum for each pair.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 14.

$$a=8 b=6$$

Rewrite -3x^{2}+14x-16 as \left(-3x^{2}+8x\right)+\left(6x-16\right).

$$\left(-3x^{2}+8x\right)+\left(6x-16\right)$$

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

$$-x\left(3x-8\right)+2\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(-x+2\right)$$

To find equation solutions, solve 3x-8=0 and -x+2=0.

$$x=\frac{8}{3} x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-2x\right)^{2}.

$$x^{2}-6x+9=25-20x+4x^{2}$$

Subtract 25 from both sides.

$$x^{2}-6x+9-25=-20x+4x^{2}$$

Subtract 25 from 9 to get -16.

$$x^{2}-6x-16=-20x+4x^{2}$$

Add 20x to both sides.

$$x^{2}-6x-16+20x=4x^{2}$$

Combine -6x and 20x to get 14x.

$$x^{2}+14x-16=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+14x-16-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+14x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 14 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-3\right)\left(-16\right)}}{2\left(-3\right)}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-3\right)\left(-16\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-14±\sqrt{196+12\left(-16\right)}}{2\left(-3\right)}$$

Multiply 12 times -16.

$$x=\frac{-14±\sqrt{196-192}}{2\left(-3\right)}$$

Add 196 to -192.

$$x=\frac{-14±\sqrt{4}}{2\left(-3\right)}$$

Take the square root of 4.

$$x=\frac{-14±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-14±2}{-6}$$

Now solve the equation x=\frac{-14±2}{-6} when ± is plus. Add -14 to 2.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

Now solve the equation x=\frac{-14±2}{-6} when ± is minus. Subtract 2 from -14.

$$x=-\frac{16}{-6}$$

Reduce the fraction \frac{-16}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

The equation is now solved.

$$x=2 x=\frac{8}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-2x\right)^{2}.

$$x^{2}-6x+9=25-20x+4x^{2}$$

Add 20x to both sides.

$$x^{2}-6x+9+20x=25+4x^{2}$$

Combine -6x and 20x to get 14x.

$$x^{2}+14x+9=25+4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+14x+9-4x^{2}=25$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+14x+9=25$$

Subtract 9 from both sides.

$$-3x^{2}+14x=25-9$$

Subtract 9 from 25 to get 16.

$$-3x^{2}+14x=16$$

Divide both sides by -3.

$$\frac{-3x^{2}+14x}{-3}=\frac{16}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{14}{-3}x=\frac{16}{-3}$$

Divide 14 by -3.

$$x^{2}-\frac{14}{3}x=\frac{16}{-3}$$

Divide 16 by -3.

$$x^{2}-\frac{14}{3}x=-\frac{16}{3}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=-\frac{16}{3}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=-\frac{16}{3}+\frac{49}{9}$$

Add -\frac{16}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{1}{3} x-\frac{7}{3}=-\frac{1}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$x=\frac{8}{3} x=2$$