To determine which of the provided choices are solutions to the given equation, let's solve the equation step by step.
Given equation:
[tex]\[ x^2 - 3x + 27 = 6x + 7 \][/tex]
Step 1: Move all terms to one side of the equation. We do this by subtracting [tex]\(6x + 7\)[/tex] from both sides:
[tex]\[ x^2 - 3x + 27 - 6x - 7 = 0 \][/tex]
Step 2: Simplify the equation:
[tex]\[ x^2 - 3x - 6x + 27 - 7 = 0 \][/tex]
[tex]\[ x^2 - 9x + 20 = 0 \][/tex]
Step 3: Factor the quadratic equation:
[tex]\[ x^2 - 9x + 20 = (x - 4)(x - 5) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex] by setting each factor to zero:
[tex]\[ x - 4 = 0 \quad \text{or} \quad x - 5 = 0 \][/tex]
[tex]\[ x = 4 \quad \text{or} \quad x = 5 \][/tex]
So, the solutions to the equation are [tex]\(x = 4\)[/tex] and [tex]\(x = 5\)[/tex].
Step 5: Check each given choice to see if it is a solution:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ 3^2 - 3(3) + 27 = 9 - 9 + 27 = 27 \][/tex]
[tex]\[ 6(3) + 7 = 18 + 7 = 25 \][/tex]
Since [tex]\(27 \neq 25\)[/tex], [tex]\(x = 3\)[/tex] is not a solution.
- For [tex]\(x = 5\)[/tex]:
[tex]\[ 5^2 - 3(5) + 27 = 25 - 15 + 27 = 37 \][/tex]
[tex]\[ 6(5) + 7 = 30 + 7 = 37 \][/tex]
Since [tex]\(37 = 37\)[/tex], [tex]\(x = 5\)[/tex] is a solution.
- For [tex]\(x = -4\)[/tex]:
[tex]\[ (-4)^2 - 3(-4) + 27 = 16 + 12 + 27 = 55 \][/tex]
[tex]\[ 6(-4) + 7 = -24 + 7 = -17 \][/tex]
Since [tex]\(55 \neq -17\)[/tex], [tex]\(x = -4\)[/tex] is not a solution.
- For [tex]\(x = 4\)[/tex]:
[tex]\[ 4^2 - 3(4) + 27 = 16 - 12 + 27 = 31 \][/tex]
[tex]\[ 6(4) + 7 = 24 + 7 = 31 \][/tex]
Since [tex]\(31 = 31\)[/tex], [tex]\(x = 4\)[/tex] is a solution.
Therefore, the solutions from the given choices that satisfy the equation are:
[tex]\[ \boxed{5} \text{ and } \boxed{4} \][/tex]
