Question 641777
Use the quadratic formula to solve for x


Note: in the case of {{{x^2-3x-40}}}, a = 1, b = -3 and c = -40


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


{{{x = (-(-3)+-sqrt((-3)^2-4(1)(-40)))/(2(1))}}}


{{{x = (3+-sqrt(9-(-160)))/(2)}}}


{{{x = (3+-sqrt(9+160))/(2)}}}


{{{x = (3+-sqrt(169))/2}}}


{{{x = (3+sqrt(169))/2}}} or {{{x = (3-sqrt(169))/2}}}


{{{x = (3+13)/2}}} or {{{x = (3-13)/2}}}


{{{x = 16/2}}} or {{{x = -10/2}}}


{{{x = 8}}}    or    {{{x = -5}}}



So the solutions are {{{x = 8}}}    or    {{{x = -5}}}