Answer:
[tex]\text{We have,}\\x^2+8x+65=0\\\text{Comparing with }ax^2+bx+c=0,\\a=1,\ b=8\text{ and }c=65\\\\\text{Using the quadratic formula,}\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-8\pm\sqrt{(-8)^2-4(1)(65)}}{2(1)}=\frac{-8\pm\sqrt{-196}}{2}=\frac{-8\pm14i}{2}\\\\\text{or, }x =-4\pm7i[/tex]
[tex]\text{Therefore, }-4+7i\text{ and }-4-7i\text{ are the roots of the given equation in}\\\text{simplest }a+bi\text{ form.}[/tex]
