Question 171875
To find the number of real solutions, simply use the discriminant formula {{{D=b^2-4ac}}}. If {{{D>0}}}, there are 2 real solutions. If {{{D=0}}}, there's only one real solutions. Finally, if {{{D<0}}}, then there are no real solutions.




From {{{x^2-10x+25}}} we can see that {{{a=1}}}, {{{b=-10}}}, and {{{c=25}}}



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



{{{D=(-10)^2-4(1)(25)}}} Plug in {{{a=1}}}, {{{b=-10}}}, and {{{c=25}}}



{{{D=100-4(1)(25)}}} Square {{{-10}}} to get {{{100}}}



{{{D=100-100}}} Multiply {{{4(1)(25)}}} to get {{{(4)(25)=100}}}



{{{D=0}}} Subtract {{{100}}} from {{{100}}} to get {{{0}}}



Since the discriminant is equal to zero, this means that there is one real solution.