Question

1. $3n^2 - 8 = 10n$

Explanation:

Step1: Rearrange the equation

We start with the equation \(3n^{2}-8 = 10n\). To solve a quadratic equation, we first write it in standard form \(ax^{2}+bx + c=0\). So, we subtract \(10n\) from both sides:
\(3n^{2}-10n - 8=0\)

Step2: Factor the quadratic (or use quadratic formula)

Let's try factoring. We need two numbers that multiply to \(3\times(- 8)=-24\) and add up to \(-10\). The numbers are \(-12\) and \(2\).
We rewrite the middle term using these numbers:
\(3n^{2}-12n + 2n-8 = 0\)
Now, factor by grouping:
\(3n(n - 4)+2(n - 4)=0\)
Factor out \((n - 4)\):
\((3n + 2)(n - 4)=0\)

Step3: Solve for n

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
Case 1: \(3n+2=0\)
Subtract \(2\) from both sides: \(3n=-2\)
Divide by \(3\): \(n=-\frac{2}{3}\)
Case 2: \(n - 4=0\)
Add \(4\) to both sides: \(n = 4\)

Answer:

\(n = 4\) or \(n=-\frac{2}{3}\)