To find the minimum value differentiate the f(x) , Use matrix calculus for differentiation (https://en.wikipedia.org/wiki/Matrix_calculus)
f'(x) =$\frac{\partial (x^{T} Ax+b^{T}+c) }{\partial x} = \frac{\partial x^{T} Ax }{\partial x} + \frac{\partial b^{T} x}{\partial x} +\frac{\partial c}{\partial x}$
$\frac{\partial x^{T} Ax }{\partial x} =(A^{T}+A)x$ (using denominator layout)
$ \frac{\partial b^{T} x}{\partial x}= b$ (using denominator layout)
$ \frac{\partial c }{\partial x} = 0$ (as c is a constant)
Hence
$ \frac{\partial x^{T} Ax }{\partial x} + \frac{\partial b^{T} x}{\partial x} +\frac{\partial c}{\partial x}= (A^{T}+A)x+b=0$ (equating f'(x) with zero for minimum)
$ (A^{T}+A)x+b =2Ax +b$ ( A is a Symmetric matrix So $ A^{T}=A$ )
hence $ x=-(A^{-1}b) /2$