Solve (3x-5)^2-8(3x-5)+8-6x=-6x+1

$$( 3 x - 5 ) ^ { 2 } - 8 ( 3 x - 5 ) + 8 - 6 x = - 6 x + 1$$

Solve for x

$$x=2$$

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-8\left(3x-5\right)+8-6x=-6x+1$$

Use the distributive property to multiply -8 by 3x-5.

$$9x^{2}-30x+25-24x+40+8-6x=-6x+1$$

Combine -30x and -24x to get -54x.

$$9x^{2}-54x+25+40+8-6x=-6x+1$$

Add 25 and 40 to get 65.

$$9x^{2}-54x+65+8-6x=-6x+1$$

Add 65 and 8 to get 73.

$$9x^{2}-54x+73-6x=-6x+1$$

Combine -54x and -6x to get -60x.

$$9x^{2}-60x+73=-6x+1$$

Add 6x to both sides.

$$9x^{2}-60x+73+6x=1$$

Combine -60x and 6x to get -54x.

$$9x^{2}-54x+73=1$$

Subtract 1 from both sides.

$$9x^{2}-54x+73-1=0$$

Subtract 1 from 73 to get 72.

$$9x^{2}-54x+72=0$$

Divide both sides by 9.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).

$$\left(x^{2}-4x\right)+\left(-2x+8\right)$$

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

$$x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-2\right)$$

To find equation solutions, solve x-4=0 and x-2=0.

$$x=4 x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-8\left(3x-5\right)+8-6x=-6x+1$$

Use the distributive property to multiply -8 by 3x-5.

$$9x^{2}-30x+25-24x+40+8-6x=-6x+1$$

Combine -30x and -24x to get -54x.

$$9x^{2}-54x+25+40+8-6x=-6x+1$$

Add 25 and 40 to get 65.

$$9x^{2}-54x+65+8-6x=-6x+1$$

Add 65 and 8 to get 73.

$$9x^{2}-54x+73-6x=-6x+1$$

Combine -54x and -6x to get -60x.

$$9x^{2}-60x+73=-6x+1$$

Add 6x to both sides.

$$9x^{2}-60x+73+6x=1$$

Combine -60x and 6x to get -54x.

$$9x^{2}-54x+73=1$$

Subtract 1 from both sides.

$$9x^{2}-54x+73-1=0$$

Subtract 1 from 73 to get 72.

$$9x^{2}-54x+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -54 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-54\right)±\sqrt{\left(-54\right)^{2}-4\times 9\times 72}}{2\times 9}$$

Square -54.

$$x=\frac{-\left(-54\right)±\sqrt{2916-4\times 9\times 72}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-54\right)±\sqrt{2916-36\times 72}}{2\times 9}$$

Multiply -36 times 72.

$$x=\frac{-\left(-54\right)±\sqrt{2916-2592}}{2\times 9}$$

Add 2916 to -2592.

$$x=\frac{-\left(-54\right)±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$x=\frac{-\left(-54\right)±18}{2\times 9}$$

The opposite of -54 is 54.

$$x=\frac{54±18}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{54±18}{18}$$

Now solve the equation x=\frac{54±18}{18} when ± is plus. Add 54 to 18.

$$x=\frac{72}{18}$$

Divide 72 by 18.

$$x=4$$

Now solve the equation x=\frac{54±18}{18} when ± is minus. Subtract 18 from 54.

$$x=\frac{36}{18}$$

Divide 36 by 18.

$$x=2$$

The equation is now solved.

$$x=4 x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-8\left(3x-5\right)+8-6x=-6x+1$$

Use the distributive property to multiply -8 by 3x-5.

$$9x^{2}-30x+25-24x+40+8-6x=-6x+1$$

Combine -30x and -24x to get -54x.

$$9x^{2}-54x+25+40+8-6x=-6x+1$$

Add 25 and 40 to get 65.

$$9x^{2}-54x+65+8-6x=-6x+1$$

Add 65 and 8 to get 73.

$$9x^{2}-54x+73-6x=-6x+1$$

Combine -54x and -6x to get -60x.

$$9x^{2}-60x+73=-6x+1$$

Add 6x to both sides.

$$9x^{2}-60x+73+6x=1$$

Combine -60x and 6x to get -54x.

$$9x^{2}-54x+73=1$$

Subtract 73 from both sides.

$$9x^{2}-54x=1-73$$

Subtract 73 from 1 to get -72.

$$9x^{2}-54x=-72$$

Divide both sides by 9.

$$\frac{9x^{2}-54x}{9}=-\frac{72}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{54}{9}\right)x=-\frac{72}{9}$$

Divide -54 by 9.

$$x^{2}-6x=-\frac{72}{9}$$

Divide -72 by 9.

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

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

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=1 x-3=-1$$

Add 3 to both sides of the equation.

$$x=4 x=2$$