Question 157986


From {{{-x^2+12x-32}}} we can see that {{{a=-1}}}, {{{b=12}}}, and {{{c=-32}}}



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



{{{D=(12)^2-4(-1)(-32)}}} Plug in {{{a=-1}}}, {{{b=12}}}, and {{{c=-32}}}



{{{D=144-4(-1)(-32)}}} Square {{{12}}} to get {{{144}}}



{{{D=144-128}}} Multiply {{{4(-1)(-32)}}} to get {{{(-4)(-32)=128}}}



{{{D=16}}} Subtract {{{128}}} from {{{144}}} to get {{{16}}}



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