Find one value of x that is a solution to the equation: {:[(3x-2)^(2)-4=9x-6],[x=◻]:}

Write and simplify equation: Write down the given equation and simplify it if possible. (3x-2)^2 - 4 = 9x - 6Expand squared term: Expand the squared term (3x-2)^2. (3x-2)(3x-2) - 4 = 9x - 6 9x^2 - 6x - 6x + 4 - 4 = 9x - 6 9x^2 - 12x = 9x - 6Move terms to one side: Move all terms to one side to set the equation to zero. 9x^2 - 12x - 9x + 6 = 0 9x^2 - 21x + 6 = 0Factor or use quadratic formula: Factor the quadratic equation if possible. Looking for two numbers that multiply to 9*6 = 54 and add up to -21 is difficult, and it seems that the quadratic does not factor nicely. We will use the quadratic formula instead. x = [-b ± sqrt(b^2 - 4ac)] / (2a) Here, a = 9, b = -21, and c = 6.Apply quadratic formula: Apply the quadratic formula. x = [-(-21) ± sqrt((-21)^2 - 4*9*6)] / (2*9) x = [21 ± sqrt(441 - 216)] / 18 x = [21 ± sqrt(225)] / 18Simplify and solve for x: Simplify under the square root and solve for x. x = [21 ± 15] / 18 We have two possible solutions: x = (21 + 15) / 18 x = 36 / 18 x = 2 or x = (21 - 15) / 18 x = 6 / 18 x = 1/3Check solutions in original equation: Check both solutions in the original equation. For x = 2: (3(2)-2)^2 - 4 = 9(2) - 6 (6-2)^2 - 4 = 18 - 6 4^2 - 4 = 12 16 - 4 = 12 12 = 12Discard incorrect solution: Since the solution x = 1/3 does not satisfy the original equation, we discard it. The solution x = 2 is the correct value that satisfies the equation.

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Solve a quadratic equation by factoring

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Q. Find one value of

x

that is a solution to the equation:

\newline

(3 x-2)^{2}-4=9 x-6

\newline

x=

Write and simplify equation: Write down the given equation and simplify it if possible. $\newline$ $(3x-2)^2 - 4 = 9x - 6$
Expand squared term: Expand the squared term $(3x-2)^2$ . $\newline$ $(3x-2)(3x-2) - 4 = 9x^2 - 6x - 6x + 4 - 4 = 9x^2 - 12x$ $\newline$ $9x^2 - 12x = 9x - 6$
Move terms to one side: Move all terms to one side to set the equation to zero. $\newline$ $9x^2 - 12x - 9x + 6 = 0$ $\newline$ $9x^2 - 21x + 6 = 0$
Factor or use quadratic formula: Factor the quadratic equation if possible. $\newline$ Looking for two numbers that multiply to $9 \times 6 = 54$ and add up to $-21$ is difficult, and it seems that the quadratic does not factor nicely. We will use the quadratic formula instead. $\newline$ $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ $\newline$ Here, $a = 9$ , $b = -21$ , and $c = 6$ .
Apply quadratic formula: Apply the quadratic formula. $\newline$ $x = \frac{{-(-21) \pm \sqrt{{(-21)^2 - 4 \cdot 9 \cdot 6}}}}{{2 \cdot 9}}$ $\newline$ $x = \frac{{21 \pm \sqrt{{441 - 216}}}}{18}$ $\newline$ $x = \frac{{21 \pm \sqrt{{225}}}}{18}$
Simplify and solve for x: Simplify under the square root and solve for x. $\newline$ $x = \frac{{21 \pm 15}}{{18}}$ $\newline$ We have two possible solutions: $\newline$ $x = \frac{{(21 + 15)}}{{18}}$ $\newline$ $x = \frac{{36}}{{18}}$ $\newline$ $x = 2$ $\newline$ or $\newline$ $x = \frac{{(21 - 15)}}{{18}}$ $\newline$ $x = \frac{{6}}{{18}}$ $\newline$ $x = \frac{1}{3}$
Check solutions in original equation: Check both solutions in the original equation. $\newline$ For $x = 2$ : $\newline$ $(3(2)-2)^2 - 4 = 9(2) - 6$ $\newline$ $(6-2)^2 - 4 = 18 - 6$ $\newline$ $4^2 - 4 = 12$ $\newline$ $16 - 4 = 12$ $\newline$ $12 = 12$
Discard incorrect solution: Since the solution $x = \frac{1}{3}$ does not satisfy the original equation, we discard it. The solution $x = 2$ is the correct value that satisfies the equation.