first of all, write it in its general form
ax^2 + bx + c = 0
t^2 - 6t + 3 = 0
a = 1, b = -6 , c = 3
t = (-b ±√(b^2-4ac) )/(2a)
= (6 ± √(36-4(1)(3))/2
= (6 ± √24)/2
= (6 ± 2√6)/2
= 3 ± √6
No complex roots here.
I am trying to solve the following using the Quadratic Formula but I am confused. Can you explain the steps?
Solve (find all complex-number solutions):
t^2 + 3 = 6t
Thank you
2 answers
Reiny,
I see it now, you took the =6t and made it a -6t for b. That was what was confusing me. Thank you so much, now I understand
I see it now, you took the =6t and made it a -6t for b. That was what was confusing me. Thank you so much, now I understand
0 answers