Question 1180506
{{{f(x) = 2x^2+3}}}
{{{g(x) = -3x+8}}}


first find {{{x}}} that makes {{{f(x) =g(x)}}}

{{{2x^2+3= -3x+8}}}

{{{2x^2+3+3x-8=0}}}

{{{2x^2+3x-5=0}}}...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-3 +- sqrt( 3^2-4*2*(-5) ))/(2*2) }}} 

{{{x = (-3 +- sqrt( 9+40 ))/4 }}} 

{{{x = (-3 +- sqrt( 49 ))/4 }}} 

{{{x = (-3 +- 7)/4 }}} 

solutions:

{{{x = (-3 + 7)/4 }}} =>{{{x=1}}}

{{{x = (-3 - 7)/4 }}} =>{{{x=-5/2}}}


calculate {{{y}}} coordinate of the intersection point


{{{f(x) = 2*1^2+3}}}
{{{f(x) = 5}}}

one  point of intersection between the following functions is: ({{{1}}},{{{5}}})

and

{{{f(x) = 2*(-5/2)^2+3}}}

{{{f(x) = 31/2}}}



second  point of intersection between the following functions is: ({{{-5/2}}},{{{31/2}}})


{{{ drawing( 600, 600, -10, 10, -10, 20, 
circle(1,5,.13),locate(1,5,p(1,5)),
circle(-5/2,31/2,.13),locate(-5/2,31/2,p(-5/2,31/2)),
graph( 600, 600, -10, 10, -10, 20, 2x^2+3, -3x+8)) }}}