Question 948989

{{{0=-3x^2+x-4}}} ....use discriminant to determine what kind of solutions you have in this case

Evaluate {{{b^2-4ac}}}
 If the value is {{{b^2-4ac>0}}} there are {{{two}}}{{{ real}}} unequal roots
If the value is {{{b^2-4ac=0}}} there is {{{one}}} real root, or  {{{two}}} equal {{{ real}}} roots
If the value is {{{b^2-4ac<0}}} there are two unequal {{{complex}}} roots

In this case you get {{{b^2-4ac=1^2-4*(-3)*(-4)=1-4*12=1-48=-47}}}

so, {{{-47<0}}} and you have two unequal {{{complex}}} roots

answer to the question how many {{{real}}} number solutions are there is {{{zero}}} or {{{none}}}