Question 24619
{{{y=x^2 + 5}}}
{{{y = 2x + 4}}}


Therefore
y=y
{{{x^2 +5 = 2x + 4}}}, which is a quadratic equation.  Solve by setting it equal to zero:
{{{x^2 + 5 - 2x - 4= 0}}}
{{{x^2 -2x + 1 = 0}}}
{{{(x-1)^2 = 0}}}
{{{x=1}}}


Substitute back into either equation, usually the simplest one is best!
y = 2x +4 
y = 2(1) + 4
y = 6


There is only one point of intersection, (1, 6) .


Just for fun, let's check by graphing:
{{{graph (400,400, -10,10,-10,10, x^2+5, 2x+4)}}}


R^2 at SCC