Respuesta :
x-intercepts are the zeros of the function.
[tex]f(x)=5(x-2)^2-3[/tex]
x-intercepts:
[tex]f(x)=0\to5(x-2)^2-3=0\ \ \ |+3\\\\5(x-2)^2=3\ \ \ |:5\\\\(x-2)^2=0.6\to x-2=\pm\sqrt{0.6}\ \ \ \ |+2\\\\x=2-\sqrt{0.6}\ \vee\ x=2+\sqrt{0.6}[/tex]
Answer: (2 - √0.6, 0) and (2 + √0.6)