Question 791496
(n+3)(n-7)=4


n^2 - 7n + 3n - 21 = 4


n^2 - 7n + 3n - 21 - 4 = 0


n^2 - 4n - 25 = 0


Now I'm going to use the quadratic formula to solve for n


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


{{{n = (-(-4)+-sqrt((-4)^2-4(1)(-25)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = -4}}}, {{{c = -25}}}


{{{n = (4+-sqrt(16-(-100)))/(2)}}}


{{{n = (4+-sqrt(16+100))/(2)}}}


{{{n = (4+-sqrt(116))/2}}}


{{{n = (4+sqrt(116))/2}}} or {{{n = (4-sqrt(116))/2}}}


{{{n = (4+2*sqrt(29))/2}}} or {{{n = (4-2*sqrt(29))/2}}}


{{{n = 2+sqrt(29)}}} or {{{n = 2-sqrt(29)}}}


{{{n = 7.385165}}} or {{{n = -3.385165}}}


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Exact Solutions: {{{n = 2+sqrt(29)}}} or {{{n = 2-sqrt(29)}}}


Approximate Solutions: {{{n = 7.385165}}} or {{{n = -3.385165}}}