Question 411559
  <pre><font size = 3 color = "indigo"><b>
Hi
Finding the solution set in a +bi form:
6x^2 + 4x +1 =0
 a=6, b=4 and c=1  |Yes!
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
x = -b  ± sqrt(b^2-4ac)]/12   |Yes! Finish substituting for b, b^2 and 4ac
x = -4 ± sqrt(16-24)]/12
x = -4 ± sqrt(-8)]/12
x = -4 ± sqrt(-1*4*2)]/12    |sqrt(-1) = i
x = -4 ± 2isqrt(2)]/12
x = -1/3 ± (1/6)isqrt(2)
x = -1/3 + (1/6)isqrt(2) x = -1/3 - (1/6)isqrt(2)