Find the solution set of the quadratic equation over the set of complex numbers. 2x2 – 6x + 7 = 0 A) x = 3 2 i or 2i B) x = 1 7 (3 − i 5 ) or 1 7 (3 + i 5 ) C) x = 1 2 (3 − i 5 ) or 1 2 (3 + i 5 ) Eliminate D) x = 1 7 (3 − i 41 ) or 1 7 (3 + i 41 )

We have the following equation:
 2x2 - 6x + 7 = 0
 Using the resolver we have:
 x = (- b +/- root (b ^ 2 - 4 * a * c)) / (2 * a)
 Substituting values we have:
 x = (- (- 6) +/- root ((- 6) ^ 2 - 4 * 2 * 7)) / (2 * 2)
 Rewriting we have:
 x = (6 +/- root (36 - 56)) / (4)
 x = (6 +/- root (-20)) / (4)
 x = (6 +/- root (-4 * 5)) / (4)
 x = (6 +/- 2raiz (-5)) / (4)
 x = (6 +/- 2raiz (-1 * 5)) / (4)
 x = (6 +/- 2raiz (5) * i) / (4)
 x = (3 +/- root (5) * i) / (2)
 The solutions are:
 x1 = (3 + root (5) * i) / (2)
 x2 = (3 - root (5) * i) / (2)
 Answer:
 x1 = (3 + root (5) * i) / (2)
 x2 = (3 - root (5) * i) / (2)