Question 180884

From {{{x^2-4x-7}}} we can see that {{{a=1}}}, {{{b=-4}}}, and {{{c=-7}}}



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



{{{D=(-4)^2-4(1)(-7)}}} Plug in {{{a=1}}}, {{{b=-4}}}, and {{{c=-7}}}



{{{D=16-4(1)(-7)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16--28}}} Multiply {{{4(1)(-7)}}} to get {{{(4)(-7)=-28}}}



{{{D=16+28}}} Rewrite {{{D=16--28}}} as {{{D=16+28}}}



{{{D=44}}} Add {{{16}}} to {{{28}}} to get {{{44}}}



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