<PRE><B>Question 209761</B>


From {{{v^2-6v-2}}} we can see that {{{a=1}}}, {{{b=-6}}}, and {{{c=-2}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(-6)^2-4(1)(-2)}}} Plug in {{{a=1}}}, {{{b=-6}}}, and {{{c=-2}}}



{{{D=36-4(1)(-2)}}} Square {{{-6}}} to get {{{36}}}



{{{D=36--8}}} Multiply {{{4(1)(-2)}}} to get {{{(4)(-2)=-8}}}



{{{D=36+8}}} Rewrite {{{D=36--8}}} as {{{D=36+8}}}



{{{D=44}}} Add {{{36}}} to {{{8}}} to get {{{44}}}



Since the discriminant is greater than zero, this means that there are two real solutions.


Since the discriminant is NOT a perfect square, this means that the square root of 44 will be irrational. So this means that you will have two different irrational (and real) solutions.


So the answer is A)</PRE>