Question 916678
{{{8x^2+6x+5=0}}}


You will use the quadratic formula to solve for x.  Recall that the quadratic formula is as follows: {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x=(-6+-sqrt(6^2-4*8*5))/(2*8)}}}----->


{{{x=(-6+-sqrt(36-160))/(16)}}}----->


{{{x=(-6+-sqrt(-124))/(16)}}}


At this point, we will be using imaginary numbers, since the number inside the square root is negative.  When we factor 124, we see that 4 x 31 = 124.  We can then reduce the root to {{{2i*sqrt(31)}}}.  We now have


{{{x=(6+-2i*sqrt(31))/(16)}}} (Ignore the ( between the +- symbol. It automatically formatted that way)


Next, we can factor out a 2 from the numerator, reducing our numerator and denominator to the following, which is our final answer:


{{{x=(-3+i*sqrt(31))/(8)}}},{{{(-3-i*sqrt(31))/(8)}}}