Let's say you have a random variable $X$, which is normally distributed according to $X \sim \mathcal{N}(1,2)$. With $1$ being the mean and $2$ being the variance. Now let's say that there is another random variable $Y$, which is defined as $Y = 10X + 5$. How do you now come up with the distribution for $Y$?
For a normal distribution, expected value equals the mean. So for $X$, the expected value is $1$. Can we now say that the $E(Y) = 10 \ E(X) + 5$, and therefore, the expected value of $Y$ is $15$?
Similarly, can we say that the variance for $Y$ is $25$?
Thus, $Y \sim \mathcal{N}(15,25)$?
The calculation of $E(Y)$ is correct. If $a$ and $b$ are constants, then $E(aX+b)=aE(X)+b$.
For the variance, use the fact that if $a$ andd $b$ are constants, then $$\text{Var}(aX+b)=a^2\text{Var}(X).\tag{1}$$
So the variance of $aX+b$ is independent of $b$. Intuitively, this is because the variance is a measure of variability, so is unchanged if you shift the distribution by a constant.
Remark: Formula (1), and generalizations, will come up quite often in applications.
Note that the expressions for $E(aX+b)$ and $\text{Var}(aX+b)$ are correct whatever distribution $X$ has.
But to get the additional conclusion that $aX+b$ has normal distribution, we need $X$ to have normal distribution.
Note that in the above calculations, we did not show that $Y$ has normal distribution. How to show it depends on the details of how the normal distribution is defined in your course. We will need to use the fact that $\Pr(Y\le y)=\Pr(10X+5\le y)=\Pr\left(X\le \frac{y-5}{10}\right)$.