Question 878981
First get everything to one side


{{{x^2=15+2x}}}


{{{x^2-2x=15}}}


{{{x^2-2x-15=0}}}


The equation is in the form {{{ax^2+bx+c=0}}} where {{{a = 1}}}, {{{b = -2}}}, {{{c = -15}}}  


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Use the quadratic formula to solve for x


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


{{{x = (-(-2)+-sqrt((-2)^2-4(1)(-15)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = -2}}}, {{{c = -15}}}  


{{{x = (2+-sqrt(4-(-60)))/(2)}}}


{{{x = (2+-sqrt(4+60))/(2)}}}


{{{x = (2+-sqrt(64))/2}}}


{{{x = (2+sqrt(64))/2}}} or {{{x = (2-sqrt(64))/2}}}


{{{x = (2+8)/2}}} or {{{x = (2-8)/2}}}


{{{x = 10/2}}} or {{{x = -6/2}}}


{{{x = 5}}}    or    {{{x = -3}}}


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The two solutions are {{{x = 5}}}    or    {{{x = -3}}}