Select the correct answer.

Which equation could be solved using this application of the quadratic formula?

[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-2)}}{2(3)} \][/tex]

A. [tex]\[ -2x^2 - 8 = 10x - 3 \][/tex]

B. [tex]\[ 3x^2 - 8x - 10 = 4 \][/tex]

C. [tex]\[ 3x^2 + 8x - 10 = -8 \][/tex]

D. [tex]\[ -2x^2 + 8x - 3 = 4 \][/tex]



Answer :

To solve this problem, we will start by comparing the given quadratic formula application to the standard form of a quadratic equation. The standard form of a quadratic equation is:

[tex]\[ ax^2 + bx + c = 0 \][/tex]

We are given the quadratic formula:

[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-2)}}{2(3)} \][/tex]

From this formula, we can deduce the coefficients:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = 8 \][/tex]
[tex]\[ c = -2 \][/tex]

Next, we need to match this to one of the given equations. Let's examine each option to see if it matches the form [tex]\( 3x^2 + 8x - 2 = 0 \)[/tex] when rearranged.

### Option A: [tex]\(-2x^2 - 8 = 10x - 3\)[/tex]
Rearranging the terms to standard form:
[tex]\[ -2x^2 - 8 = 10x - 3 \\ -2x^2 - 10x - 8 + 3 = 0 \\ -2x^2 - 10x - 5 = 0 \][/tex]
This does not match the form [tex]\( 3x^2 + 8x - 2 = 0 \)[/tex].

### Option B: [tex]\(3x^2 - 8x - 10 = 4\)[/tex]
Rearranging the terms to standard form:
[tex]\[ 3x^2 - 8x - 10 = 4 \\ 3x^2 - 8x - 10 - 4 = 0 \\ 3x^2 - 8x - 14 = 0 \][/tex]
This does not match the form [tex]\( 3x^2 + 8x - 2 = 0 \)[/tex].

### Option C: [tex]\(3x^2 + 8x - 10 = -8\)[/tex]
Rearranging the terms to standard form:
[tex]\[ 3x^2 + 8x - 10 = -8 \\ 3x^2 + 8x - 10 + 8 = 0 \\ 3x^2 + 8x - 2 = 0 \][/tex]
This matches the form [tex]\( 3x^2 + 8x - 2 = 0 \)[/tex].

### Option D: [tex]\(-2x^2 + 8x - 3 = 4\)[/tex]
Rearranging the terms to standard form:
[tex]\[ -2x^2 + 8x - 3 = 4 \\ -2x^2 + 8x - 3 - 4 = 0 \\ -2x^2 + 8x - 7 = 0 \][/tex]
This does not match the form [tex]\( 3x^2 + 8x - 2 = 0 \)[/tex].

Thus, the only equation that matches the given standard form is option:

[tex]\[ C: 3x^2 + 8x - 10 = -8 \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{C} \][/tex]