Sure! Let's solve the given quadratic equation step-by-step.
We start with:
[tex]\[ 2n^2 = 10n + 48 \][/tex]
First, we need to bring all terms to one side of the equation to set it to zero:
[tex]\[ 2n^2 - 10n - 48 = 0 \][/tex]
Now, we have a standard form quadratic equation [tex]\( an^2 + bn + c = 0 \)[/tex] where [tex]\( a = 2 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = -48 \)[/tex].
To solve this quadratic equation, we can use factoring or the quadratic formula. The quadratic formula is:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\( a = 2 \)[/tex]
[tex]\( b = -10 \)[/tex]
[tex]\( c = -48 \)[/tex]
Now, we plug these values into the quadratic formula:
[tex]\[ n = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 2 \cdot (-48)}}{2 \cdot 2} \][/tex]
Simplify inside the square root:
[tex]\[ n = \frac{10 \pm \sqrt{100 + 384}}{4} \][/tex]
[tex]\[ n = \frac{10 \pm \sqrt{484}}{4} \][/tex]
[tex]\[ n = \frac{10 \pm 22}{4} \][/tex]
This gives us two solutions:
[tex]\[ n = \frac{10 + 22}{4} = \frac{32}{4} = 8 \][/tex]
[tex]\[ n = \frac{10 - 22}{4} = \frac{-12}{4} = -3 \][/tex]
Therefore, the values for [tex]\( n \)[/tex] are:
[tex]\[ n = 8 \quad \text{or} \quad n = -3 \][/tex]
The correct answer is:
D. [tex]\(-3\)[/tex] or [tex]\(8\)[/tex]
