Question 152559
Are you trying to solve this as a system? If you are, then there are two points that satisfy this system


{{{2x+2y=4 }}} Start with the first equation.



{{{2x+2(x^2-4)=4 }}} Plug in {{{y=x^2-4}}}



{{{2x+2x^2-8=4 }}} Distribute



{{{2x+2x^2-8-4=0}}} Get all terms to the left side.



{{{2x^2+2x-12=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=2}}}, and {{{c=-12}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(2) +- sqrt( (2)^2-4(2)(-12) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=2}}}, and {{{c=-12}}}



{{{x = (-2 +- sqrt( 4-4(2)(-12) ))/(2(2))}}} Square {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- sqrt( 4--96 ))/(2(2))}}} Multiply {{{4(2)(-12)}}} to get {{{-96}}}



{{{x = (-2 +- sqrt( 4+96 ))/(2(2))}}} Rewrite {{{sqrt(4--96)}}} as {{{sqrt(4+96)}}}



{{{x = (-2 +- sqrt( 100 ))/(2(2))}}} Add {{{4}}} to {{{96}}} to get {{{100}}}



{{{x = (-2 +- sqrt( 100 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- 10)/(4)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{x = (-2 + 10)/(4)}}} or {{{x = (-2 - 10)/(4)}}} Break up the expression. 



{{{x = (8)/(4)}}} or {{{x =  (-12)/(4)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -3}}} Simplify. 



So the answers are {{{x = 2}}} or {{{x = -3}}} 



So the two solutions are (2,0) and (-3,5)