Let the quadratic equation is

Here, a is the coefficient of x^2, b is the coefficient of x and c is the constant.
For the equation we have a = 1, b = -8 and c = 16.
We know that the quadratic formula is given by:
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
So, the solution of the quadratic equation is:
![\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(16)}}{2(1)} \\ x=\frac{8\pm\sqrt[]{64-64}}{2} \\ x=\frac{8\pm\sqrt[]{0}}{2} \\ x=(8)/(2) \\ x=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/frd7g1j52o45rwbipd7ml83wsmc9nb6gu9.png)
Thus, there are two real and equal solutions for the given quadratic equation that is x = 4 and x = 4.