Question 727187
{{{y=x^2+2x-3}}}...eq. 1...parabola
{{{y=5}}}...eq. 2.......horizontal line
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{{{y=x^2+2x-3}}}..substitute {{{y}}} from eq. 2

{{{5=x^2+2x-3}}}....solve for {{{x}}}

{{{0=x^2+2x-3-5}}}

{{{x^2+2x-8=0}}}..use quadratic formula

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

{{{x = (-2 +- sqrt( 2^2-4*1*(-8) ))/(2*1) }}} 

{{{x = (-2 +- sqrt( 4+32 ))/2 }}} 

{{{x = (-2 +- sqrt(36 ))/2 }}} 

{{{x = (-2 +- 6)/2 }}} 

solutions:

{{{x = (-2 + 6)/2 }}} 

{{{x = 4/2 }}}

{{{x = 2 }}}

or

{{{x = (-2- 6)/2 }}} 

{{{x = -8/2 }}}

{{{x = -4}}}

so, parabola and horizontal line intersecting in two points and they are:({{{2}}},{{{5}}}) and ({{{-4}}},{{{5}}})

{{{drawing( 600, 600, -10, 10, -10, 10,circle(2,5,0.2),circle(-4,5,0.2), graph( 600, 600, -10, 10, -10, 10, x^2+2x-3, 5)) }}}