To solve the equations you've provided, I'll take them one at a time.

**1. For the first equation:**

\[ 2x^2 + 6 = 102 \]

First, let's isolate \( x^2 \):

1. Subtract 6 from both sides:
\[
2x^2 = 102 - 6
\]
\[
2x^2 = 96
\]

2. Now, divide both sides by 2:
\[
x^2 = \frac{96}{2}
\]
\[
x^2 = 48
\]

3. Now take the square root of both sides:
\[
x = \pm \sqrt{48}
\]
\[
x = \pm \sqrt{16 \cdot 3} = \pm 4\sqrt{3}
\]

So the solutions to the first equation are:
\[ x = 4\sqrt{3} \quad \text{and} \quad x = -4\sqrt{3} \]

---

**2. For the second equation:**

\[ x^2 + 6 = 10 \]

Again, isolate \( x^2 \):

1. Subtract 6 from both sides:
\[
x^2 = 10 - 6
\]
\[
x^2 = 4
\]

2. Take the square root of both sides:
\[
x = \pm \sqrt{4}
\]
\[
x = \pm 2
\]

So the solutions to the second equation are:
\[ x = 2 \quad \text{and} \quad x = -2 \]

---

In summary, the solutions are:
1. For \( 2x^2 + 6 = 102 \): \( x = 4\sqrt{3} \) and \( x = -4\sqrt{3} \)
2. For \( x^2 + 6 = 10 \): \( x = 2 \) and \( x = -2 \)