SOLUTION: -6 2/3 x2 + 1 1/3 x = 2/3 A) x = (-1 +- 10i)/3 B) x = (-1 +- 3i)/10 C) x = (1 +- 3i)/-10 D) x = (-1 +- 3i)/-10 E) x = (-1 +- 10i)/-3 F) x = (2 +- 3i)/6 10. -57 3/5 x2 + 3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: -6 2/3 x2 + 1 1/3 x = 2/3 A) x = (-1 +- 10i)/3 B) x = (-1 +- 3i)/10 C) x = (1 +- 3i)/-10 D) x = (-1 +- 3i)/-10 E) x = (-1 +- 10i)/-3 F) x = (2 +- 3i)/6 10. -57 3/5 x2 + 3       Log On


   

Question 181480: -6 2/3 x2 + 1 1/3 x = 2/3
A) x = (-1 +- 10i)/3
B) x = (-1 +- 3i)/10
C) x = (1 +- 3i)/-10
D) x = (-1 +- 3i)/-10
E) x = (-1 +- 10i)/-3
F) x = (2 +- 3i)/6
10. -57 3/5 x2 + 3 3/5 x = -39 3/5 x2 - 7 1/5 x + 14 2/5
A) x = {2 +- i[SQRT(75)}/10
B) x = {3 +- i[SQRT(71)]}/-10
C) x = {-3 +- i[SQRT(57)]}/7
D) x = {-3 +- i[SQRT(71)]}/-10
E) x = {-2 +- i[SQRT(75)]}/-7
F) x = {2 +- i[SQRT(39)]}/14
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The best approach to both of these is to get rid of the fractional coefficients. For instance,



Start by converting the mixed fractions to improper fractions:



Multiply both sides by -3:



Add 2 to both sides:



Multiply both sides by



Now you have an equivalent equation with integer factors that can be easily solved with the quadratic formula.

Follow a similar process for your second problem.

John