Question 755522
x^2 + (x+2)^2 = 6500


x^2 + x^2 + 4x + 4 = 6500


2x^2 + 4x + 4 = 6500


2x^2 + 4x + 4 - 6500 = 0


2x^2 + 4x - 6496 = 0



Use the quadratic formula to solve for x


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


{{{x = (-(4)+-sqrt((4)^2-4(2)(-6496)))/(2(2))}}} Plug in {{{a = 2}}}, {{{b = 4}}}, {{{c = -6496}}}


{{{x = (-4+-sqrt(16-(-51968)))/(4)}}}


{{{x = (-4+-sqrt(16+51968))/(4)}}}


{{{x = (-4+-sqrt(51984))/4}}}


{{{x = (-4+sqrt(51984))/4}}} or {{{x = (-4-sqrt(51984))/4}}}


{{{x = (-4+228)/4}}} or {{{x = (-4-228)/4}}}


{{{x = 224/4}}} or {{{x = -232/4}}}


{{{x = 56}}} or {{{x = -58}}}


x = 56


x + 2 = 56 + 2 = 58


So the two consecutive positive even integers are 56 and 58.