Answer:
[tex]x_{1,2}=4,\ 12[/tex]
Step-by-step explanation:
Consider the polynmial function [tex]f(x)=x^2-16x+48.[/tex] The zeros of the polynomial function are the values of x at which f(x)=0. Solve the equation [tex]x^2-16x+48=0:[/tex]
[tex]D=b^2-4ac=(-16)^2-4\cdot 1\cdot 48=256-192=64=8^2,\\ \\x_{1,2}=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-(-16)\pm \sqrt{64}}{2\cdot 1}=\dfrac{16\pm 8}{2}=4,\ 12.[/tex]
