Calculate Hessian of a "weird" function

let be $L $ an invertible matrix, $b in mathbb{R}^{n}$ and $mu in mathbb{R}$

and $J: mathbb{R}^{n} rightarrow mathbb{R}$ defined as $J(x)=|L x|_{2}^{2}+mu x^{t} x-b^{t} x$

How do I calculate its hessian matrix?
I cant operate as a usual function and calculate its gradient and also i dont know what to do with that norm

Progress:

$|L x|_{2}^{2} =sum_{i=1}^{n}sum_{j=1}^{n} (l_{i,j}cdot x_{j})^2$

$mu x^{t}cdot x =sum_{i=1}^{n} mu cdot x_{i}^2$

$-b^{t} x = – sum_{i=1}^{n} b_{i} cdot x_{i}$

so $J(x)=\sum_{i=1}^{n}sum_{j=1}^{n} (l_{i,j}cdot x_{j})^2+sum_{i=1}^{n} mu cdot x_{i}^2 – sum_{i=1}^{n} b_{i} cdot x_{i} $

so the gradient is

$nabla J(x)=(sum_{i=1}^{n} l_{i,1} + 2 cdot mu cdot x_{1} – b_{1}, …, sum_{i=1}^{n} l_{i,n} + 2 cdot mu cdot x_{n} – b_{n} ) $

and the Hessian is the diagonal matrix of $2 cdot mu$

Dunno if it is right and I also think there has to be an easier way of doing it, like operating with the vectors and matrix instead of deriving each variable on their own.

FINAL EDIT:

As some comments sugested(thank you all) i can write the function as :
$J(x)=x^tL^tLx + mu x^tx -b^tx$

So:

$ nabla J(x)= 2(L^tL+Imu)x-b$

and

$HJ(x)=2LL^t+2 I mu$