Let \(p\) be an odd prime, and \(a\), \(b\), \(q\) integers such that \((a,b,p)=1\) and \((q,p)=1\). Suppose that \(\tau\) is the order of \(q\bmod p\), \(\Delta= \text{ord}_ p (q^ \tau-1)\), \(\alpha\geq\Delta\) is an integer, \(\tau_ \alpha= p^{\alpha-\Delta}\tau\), \(\varepsilon_ 0= q^ \tau\) and \(U_ p\) is the group of the units of the field \(\mathbb{Q}_ p\). Furthermore, \(\hat f_ r(X)= aq^ r p^{-\Delta} (\varepsilon_ 0^ x-1)+ bX\) \((0\leq r\leq\tau-1)\) is a polynomial over \(\mathbb{Z}_ p\), \(\varepsilon\in U_ p\), \(r_ 0\) is an integer such that the inequality \(\text{ord}_ p (aq^ r \varepsilon+b)\geq \text{ord}_ p(ap^ \Delta)\) is satisfied, \(\theta\in \mathbb{Z}_ p\) is an integral \(p\)-adic zero of the derivative \(\hat f_ r'(X)\). The object of this paper is the determination of the exact value of the sum \[ S(a,q;b,p^ \alpha)= \sum_{x=1}^{\tau_ \alpha} \exp(2\pi i (aq^ x/ p^ \alpha+ bx/ \tau_ \alpha)) \] with the aid of the method of \(p\)-adic isometries. It is shown that for all \(\alpha> \Delta+2e\) the equality \[ S(a,q;b,p)= \exp(2\pi i(aq^{r_ 0}/ p^ \alpha+ br_ 0/ \tau_ \alpha+ \hat f_{r_ 0} (\theta)/ p^{\alpha+\Delta}))\times SG_ 2(\hat f_{r_ 0}^{\prime\prime} (\theta)/ 2!, p^{\alpha- \Delta}) \] holds, where \(SG_ 2(A: p^ \beta)= \sum_{x=1}^{p^ \beta} \exp(2\pi iAX^ 2/ p^ \beta)\) is the quadratic Gauss sum.