Question 1074569
DISCLAIMER: You probably want someone to check your work,
but I make arithmetic mistakes too, so if our solutions disagree,
it just may be that yours is right.

The medians are the lines connecting one vertex to the midpoint of the other side.
So, we need to connect point {{{A(-3,4)}}} to point {{{M}}}, the midpoint of {{{BC}}} ,
and we need to connect point {{{B(5.8)}}} to point {{{N}}}, the midpoint of {{{AC}}} .
 
Finding midpoints:
The coordinates of a midpoint are the averages of the coordinates of the endpoints.
{{{x[M]=(5+6)/2=11/2=5.5}}} , and
{{{y[M]=(8-3)/2=5/2=2.5}}} , giving us {{{M(5.5,2.5)}}} .
{{{x[N]=(-3+6)/2=3/2=1.5}}} , and
{{{y[N]=(4-3)/2=1/2=0.5}}} , giving us {{{N(1.5,0.5)}}} .
 
Finding the equation for line AM:
The slope is {{{(2.5-4)/(5.5-(-3))=-1.5/(5.5+3)=-1.5/8.5=-3/17}}} .
The equation for line AM, in {{{highlight(point-slope)}}} form, using point {{{A(-3,4)}}} is
{{{y-4=(-3/17)*(x-(-3))}}} --> {{{highlight(y-4=-(3/17)(x+3))}}} .
If we want the more traditional and unique {{{highlight(slope-intercept)}}} form, we do some more algebra
{{{y-4=-(3/17)(x+3)}}} --> {{{y-4=-(3/17)x-9/17}}} --> {{{y=-(3/17)x-9/17+4}}} --> {{{highlight(y=-(3/17)x+59/17)}}}
 
Finding the equation for line BN:
The slope is {{{(8-0.5)/(5-1.5)=7.5/3.5=15/7}}} .
The equation for line BN, in {{{highlight(point-slope)}}} form, using point {{{B(5,8)}}} is
{{{highlight(y-8=(15/7)(x-5))}}} .
If we want the more traditional and unique {{{highlight(slope-intercept)}}} form, we do some more algebra
{{{y-8=(15/7)(x-5)}}} --> {{{y-8=(15/7)x-75/7}}} --> {{{y=(15/7)x-75/7+8}}} --> {{{highlight(y=(15/7)x-19/17)}}}