The formatting seems to have taken a hit there, so we'll assume that the equations are

#y=x^2-6x+23# and

#y-5=3x#.

The second equation can be written as #y=3x+5#. Both equations need to be true for specific values of #x# and #y#, so we can substitute the value of #y# in terms of #x# in the first equation to solve for #x#:

#3x+5=x^2-6x+23=>x^2-9x+18=0#

This is just a quadratic equation, so let's find the discriminant:

#D=(-9)^2-4*1*18=81-72=9>0#

Since the discriminant is greater than #0#, we know the equation has two distinct roots.

#x_1=(9+3)/2=6#

#x_2=(9-3)/2=3#

Then we just need to use those values of #x# to determine #y# using #y=3x+5#. If #x=6#, #y=23# and if #x=3#, #y=14#

Therefore, the graphs of the two equations intersect at the points #(6,23)# and #(3,14)#

NOTE: For any equation of the form #ax^2+bx+c=0#, the discriminant is calculated by #D=b^2-4ac# and the roots are calculated by #(-b+-sqrtD)/(2a)# if #D>=0#, and if #D<0# there are no real roots.