```Question 187799
To perform elimination, simply subtract equation 2 from equation 1 to get

{{{y-y=(x^2 + 5x + 6)-(x+11)}}}

{{{0=x^2 + 5x + 6-x-11}}}

{{{0=x^2+4x-5}}}

Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=4}}}, and {{{c=-5}}}

Let's use the quadratic formula to solve for x

{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula

{{{x = (-(4) +- sqrt( (4)^2-4(1)(-5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=4}}}, and {{{c=-5}}}

{{{x = (-4 +- sqrt( 16-4(1)(-5) ))/(2(1))}}} Square {{{4}}} to get {{{16}}}.

{{{x = (-4 +- sqrt( 16--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}

{{{x = (-4 +- sqrt( 16+20 ))/(2(1))}}} Rewrite {{{sqrt(16--20)}}} as {{{sqrt(16+20)}}}

{{{x = (-4 +- sqrt( 36 ))/(2(1))}}} Add {{{16}}} to {{{20}}} to get {{{36}}}

{{{x = (-4 +- sqrt( 36 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.

{{{x = (-4 +- 6)/(2)}}} Take the square root of {{{36}}} to get {{{6}}}.

{{{x = (-4 + 6)/(2)}}} or {{{x = (-4 - 6)/(2)}}} Break up the expression.

{{{x = (2)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms.

{{{x = 1}}} or {{{x = -5}}} Simplify.

So the answers are {{{x = 1}}} or {{{x = -5}}}

Now plug these values back into either equation (I'm going to use the second)

Plug in {{{x=1}}}

{{{y=x+11=1+11=12}}}

So the first intersection is (1,12)

----------------------

Plug in {{{x=-5}}}

{{{y=x+11=-5+11=6}}}

So the second intersection is (-5,6)

=================================================================

So the solutions are

{{{x=1}}} and {{{y=12}}} which gives the ordered pair (1,12)

OR

{{{x=-5}}} and {{{y=6}}} which gives the ordered pair (-5,6)

Here's the graph to visually confirm the answer:

{{{ drawing(500, 500, -10, 10, -7, 13,
grid(1),
graph( 500, 500, -10, 10, -7, 13,x^2 + 5x + 6,x+11)

)}}} Graph of {{{y = x^2 + 5x + 6}}} and {{{y=x+11}}}```