Question 392768
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Hi
Given x = -4 or 6, those are the 'roots' of the quadratic equation and 
Note: {{{(x - r[1])(x - r[2])= y }}}
  y = (x - (-4))(x-6)
  y = (x+4)(x-6)     |multiplying gives us
  y = x^2 + 4x - 6x - 24
  y = x^2 - 2x - 24 = 0
therefore, in this example of a standard quadratic equation y = x^2 + bx + c
 b = -2 and c = -24