Question 36186
As I posted earlier on a similar problem, just calculate {{{b^2 - 4ac}}}, where in this case, {{{2x^2 - 10x + 25 = 0}}}, a =2, b= -10, and c= 25.


{{{b^2 - 4ac}}}
{{{(-10)^2 - 4*2*25}}}
{{{100-200}}}
{{{-100}}}


Since this discriminant is less than 0, there will be NO REAL solutions.  The solutions will be complex.  They will also always be conjugate pairs--no extra charge!


R^2 at SCC