Question 941422

{{{y = x^2 + 3x + 9}}} =>In this case {{{a=1}}},{{{b=3}}},{{{c=9}}}

The discriminant is {{{b^2-4ac =3^2-4*3*9=9-108=-99}}}:

If the value is {{{b^2-4ac >0}}} and a Perfect Square there are Two Real Rational Solutions

If the value is {{{b^2-4ac >0}}} and is not a Perfect Square there are Two Real Irrational Solutions

If the value is {{{b^2-4ac=0}}} there is One Real Solution

If the value is {{{b^2-4ac<0}}} there are No Real Solutions, there are Two Imaginary Solutions

in your case {{{b^2-4ac=-99}}} which is < then {{{0}}};

so, answer is C. No real roots