Solve (x-3)^2=7(x-3)

$$( x - 3 ) ^ { 2 } = 7 ( x - 3 )$$

Solve for x

$$x=10$$

$$x=3$$

Steps Using Factoring

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

$$x^{2}-6x+9=7\left(x-3\right)$$

Use the distributive property to multiply 7 by x-3.

$$x^{2}-6x+9=7x-21$$

Subtract 7x from both sides.

$$x^{2}-6x+9-7x=-21$$

Combine -6x and -7x to get -13x.

$$x^{2}-13x+9=-21$$

Add 21 to both sides.

$$x^{2}-13x+9+21=0$$

Add 9 and 21 to get 30.

$$x^{2}-13x+30=0$$

To solve the equation, factor x^{2}-13x+30 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x-3\right)$$

To find equation solutions, solve x-10=0 and x-3=0.

$$x=10 x=3$$

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}.

$$x^{2}-6x+9=7\left(x-3\right)$$

Use the distributive property to multiply 7 by x-3.

$$x^{2}-6x+9=7x-21$$

Subtract 7x from both sides.

$$x^{2}-6x+9-7x=-21$$

Combine -6x and -7x to get -13x.

$$x^{2}-13x+9=-21$$

Add 21 to both sides.

$$x^{2}-13x+9+21=0$$

Add 9 and 21 to get 30.

$$x^{2}-13x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 30=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite x^{2}-13x+30 as \left(x^{2}-10x\right)+\left(-3x+30\right).

$$\left(x^{2}-10x\right)+\left(-3x+30\right)$$

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

$$x\left(x-10\right)-3\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-3\right)$$

To find equation solutions, solve x-10=0 and x-3=0.

$$x=10 x=3$$

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}.

$$x^{2}-6x+9=7\left(x-3\right)$$

Use the distributive property to multiply 7 by x-3.

$$x^{2}-6x+9=7x-21$$

Subtract 7x from both sides.

$$x^{2}-6x+9-7x=-21$$

Combine -6x and -7x to get -13x.

$$x^{2}-13x+9=-21$$

Add 21 to both sides.

$$x^{2}-13x+9+21=0$$

Add 9 and 21 to get 30.

$$x^{2}-13x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 30}}{2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 30}}{2}$$

Multiply -4 times 30.

$$x=\frac{-\left(-13\right)±\sqrt{169-120}}{2}$$

Add 169 to -120.

$$x=\frac{-\left(-13\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-13\right)±7}{2}$$

The opposite of -13 is 13.

$$x=\frac{13±7}{2}$$

Now solve the equation x=\frac{13±7}{2} when ± is plus. Add 13 to 7.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{13±7}{2} when ± is minus. Subtract 7 from 13.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=10 x=3$$

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}.

$$x^{2}-6x+9=7\left(x-3\right)$$

Use the distributive property to multiply 7 by x-3.

$$x^{2}-6x+9=7x-21$$

Subtract 7x from both sides.

$$x^{2}-6x+9-7x=-21$$

Combine -6x and -7x to get -13x.

$$x^{2}-13x+9=-21$$

Subtract 9 from both sides.

$$x^{2}-13x=-21-9$$

Subtract 9 from -21 to get -30.

$$x^{2}-13x=-30$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-30+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=-30+\frac{169}{4}$$

Add -30 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{49}{4}$$

Factor x^{2}-13x+\frac{169}{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}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{7}{2} x-\frac{13}{2}=-\frac{7}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$x=10 x=3$$