Question 387441
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Hi,   
 Yes, when b^2 - 4ac < 0 in the quadratic equaton ax^2 + bx + c = 0 
that will result in {{{sqrt(b^2 - 4ac )}}} being an imaginary number.
solving using the quadratic equation
t^2 - 6t + 10 = 0
  {{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
  {{{t = (6 +- sqrt( -4))/(2) }}}
  {{{t = (6 +- 2i)/(2) }}}
  {{{t = 3 +- i) }}}
this quadraic equation has no real-number solutions