Question

1. $9n^2 - 6 = -164$
2. $-9 - 9n^2 = -909$

Explanation:

Step1: Solve equation 13: Isolate $n^2$

Add 6 to both sides:
$9n^2 - 6 + 6 = -164 + 6$
$9n^2 = -158$
Divide by 9:
$n^2 = \frac{-158}{9}$
Since the square of a real number cannot be negative, there are no real solutions for equation 13.

Step2: Solve equation 14: Isolate $n^2$

Add 9 to both sides:
$-9 - 9n^2 + 9 = -909 + 9$
$-9n^2 = -900$
Divide by -9:
$n^2 = 100$
Take square root of both sides:
$n = \pm\sqrt{100}$
$n = \pm10$

Answer:

1. No real solutions
2. $n = 10$ or $n = -10$