Question 1085503
{{{x^2+75=0}}}
.
.
.
{{{a=1}}}
{{{b=0}}}
{{{c=75}}}
.
.
.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-0 +- sqrt( 0^2-4*1*75 ))/(2*1) }}} 
{{{x = (0 +- sqrt( -300 ))/(2) }}} 
Since the value under the square root is negative, there are no real roots to this equation only complex ones.