How do you solve #-6x ^ { 2} = 3x + 2#?

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Answer 1 · 2016-12-22T15:22:03

There are no real solutions for x. The irrational solutions are:
#x=frac{-3+-isqrt(-39)}{12}#

#-6x^2=3x+2#

To solve the quadratic equation, get zero on one side by adding #6x^2# to each side:

#0=6x^2+3x+2#

Use the quadratic formula to solve: #x=frac{-b+-sqrt(b^2-4ac)}{2a}#

#x=frac{-3+-sqrt(3^2-4(6)(2))}{2(6)}#

#x=frac{-3+-sqrt(-39)}{12}#

#x=frac{-3+-isqrt(-39)}{12}#