Question 144828

{{{D=b^2-4ac}}} Start with the discriminant formula



From {{{3z^2+z-1}}} we can see that {{{a=3}}}, {{{b=1}}}, and {{{c=-1}}}



{{{D=(1)^2-4(3)(-1)}}} Plug in {{{a=3}}}, {{{b=1}}}, and {{{c=-1}}}



{{{D=1-4(3)(-1)}}} Square {{{1}}} to get {{{1}}}



{{{D=1--12}}} Multiply {{{4(3)(-1)}}} to get {{{(12)(-1)=-12}}}



{{{D=1+12}}} Rewrite {{{D=1--12}}} as {{{D=1+12}}}



{{{D=13}}} Add {{{1}}} to {{{12}}} to get {{{13}}}



Since the discriminant is greater than zero, this means that there are two real solutions