Question 291123
You're on the right track. I'll start where you left off.



Let's solve {{{ 13x^2+65x+78=0 }}} by use of the quadratic formula.


*[invoke quadratic_formula 13, 65, 78, "x"]



Now plug each of these solutions into {{{y=5x+13}}} to find the corresponding 'y' coordinates of the intersections. 


So when {{{x=-2}}}, then {{{y=5x+13=5(-2)+13=-10+13=3}}} giving us the point (-2,3)



Also, when {{{x=-3}}}, then {{{y=5x+13=5(-3)+13=-15+13=-2}}} giving us the  point (-3,-2)



So the two points of intersection are (-2,3) and (-3,-2)