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

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

Solve for x

$$x=-3$$

$$x = \frac{24}{7} = 3\frac{3}{7} \approx 3.428571429$$

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}.

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

Consider \left(3x-8\right)\left(3x+8\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 8.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Expand \left(3x\right)^{2}.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

Subtract 64 from 9 to get -55.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

Add -55 and 1 to get -54.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

Use the distributive property to multiply x by x+3.

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

Use the distributive property to multiply 3 by x^{2}+3x+6.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Subtract 3x^{2} from both sides.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+6x-54=9x+18$$

Subtract 9x from both sides.

$$7x^{2}+6x-54-9x=18$$

Combine 6x and -9x to get -3x.

$$7x^{2}-3x-54=18$$

Subtract 18 from both sides.

$$7x^{2}-3x-54-18=0$$

Subtract 18 from -54 to get -72.

$$7x^{2}-3x-72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-72. To find a and b, set up a system to be solved.

$$a+b=-3 ab=7\left(-72\right)=-504$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -504.

$$1,-504 2,-252 3,-168 4,-126 6,-84 7,-72 8,-63 9,-56 12,-42 14,-36 18,-28 21,-24$$

Calculate the sum for each pair.

$$1-504=-503 2-252=-250 3-168=-165 4-126=-122 6-84=-78 7-72=-65 8-63=-55 9-56=-47 12-42=-30 14-36=-22 18-28=-10 21-24=-3$$

The solution is the pair that gives sum -3.

$$a=-24 b=21$$

Rewrite 7x^{2}-3x-72 as \left(7x^{2}-24x\right)+\left(21x-72\right).

$$\left(7x^{2}-24x\right)+\left(21x-72\right)$$

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

$$x\left(7x-24\right)+3\left(7x-24\right)$$

Factor out common term 7x-24 by using distributive property.

$$\left(7x-24\right)\left(x+3\right)$$

To find equation solutions, solve 7x-24=0 and x+3=0.

$$x=\frac{24}{7} x=-3$$

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}.

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

Consider \left(3x-8\right)\left(3x+8\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 8.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Expand \left(3x\right)^{2}.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

Subtract 64 from 9 to get -55.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

Add -55 and 1 to get -54.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

Use the distributive property to multiply x by x+3.

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

Use the distributive property to multiply 3 by x^{2}+3x+6.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Subtract 3x^{2} from both sides.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+6x-54=9x+18$$

Subtract 9x from both sides.

$$7x^{2}+6x-54-9x=18$$

Combine 6x and -9x to get -3x.

$$7x^{2}-3x-54=18$$

Subtract 18 from both sides.

$$7x^{2}-3x-54-18=0$$

Subtract 18 from -54 to get -72.

$$7x^{2}-3x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -3 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 7\left(-72\right)}}{2\times 7}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 7\left(-72\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-3\right)±\sqrt{9-28\left(-72\right)}}{2\times 7}$$

Multiply -28 times -72.

$$x=\frac{-\left(-3\right)±\sqrt{9+2016}}{2\times 7}$$

Add 9 to 2016.

$$x=\frac{-\left(-3\right)±\sqrt{2025}}{2\times 7}$$

Take the square root of 2025.

$$x=\frac{-\left(-3\right)±45}{2\times 7}$$

The opposite of -3 is 3.

$$x=\frac{3±45}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{3±45}{14}$$

Now solve the equation x=\frac{3±45}{14} when ± is plus. Add 3 to 45.

$$x=\frac{48}{14}$$

Reduce the fraction \frac{48}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{24}{7}$$

Now solve the equation x=\frac{3±45}{14} when ± is minus. Subtract 45 from 3.

$$x=-\frac{42}{14}$$

Divide -42 by 14.

$$x=-3$$

The equation is now solved.

$$x=\frac{24}{7} x=-3$$

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}.

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

Consider \left(3x-8\right)\left(3x+8\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 8.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Expand \left(3x\right)^{2}.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

Subtract 64 from 9 to get -55.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

Add -55 and 1 to get -54.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

Use the distributive property to multiply x by x+3.

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

Use the distributive property to multiply 3 by x^{2}+3x+6.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Subtract 3x^{2} from both sides.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+6x-54=9x+18$$

Subtract 9x from both sides.

$$7x^{2}+6x-54-9x=18$$

Combine 6x and -9x to get -3x.

$$7x^{2}-3x-54=18$$

Add 54 to both sides.

$$7x^{2}-3x=18+54$$

Add 18 and 54 to get 72.

$$7x^{2}-3x=72$$

Divide both sides by 7.

$$\frac{7x^{2}-3x}{7}=\frac{72}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{3}{7}x=\frac{72}{7}$$

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

$$x^{2}-\frac{3}{7}x+\left(-\frac{3}{14}\right)^{2}=\frac{72}{7}+\left(-\frac{3}{14}\right)^{2}$$

Square -\frac{3}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{7}x+\frac{9}{196}=\frac{72}{7}+\frac{9}{196}$$

Add \frac{72}{7} to \frac{9}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{7}x+\frac{9}{196}=\frac{2025}{196}$$

Factor x^{2}-\frac{3}{7}x+\frac{9}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{14}\right)^{2}=\frac{2025}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{14}\right)^{2}}=\sqrt{\frac{2025}{196}}$$

Simplify.

$$x-\frac{3}{14}=\frac{45}{14} x-\frac{3}{14}=-\frac{45}{14}$$

Add \frac{3}{14} to both sides of the equation.

$$x=\frac{24}{7} x=-3$$