Question 878956
I'm going to refer to {{{a^2-7a-10=0}}} as {{{x^2-7x-10=0}}}


{{{x^2-7x-10=0}}} is in the form {{{ax^2+bx+c=0}}} where {{{a = 1}}}, {{{b = -7}}}, {{{c = -10}}}  


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Use the quadratic formula to solve for x


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


{{{x = (-(-7)+-sqrt((-7)^2-4(1)(-10)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = -7}}}, {{{c = -10}}}  


{{{x = (7+-sqrt(49-(-40)))/(2)}}}


{{{x = (7+-sqrt(49+40))/(2)}}}


{{{x = (7+-sqrt(89))/2}}}


{{{x = (7+sqrt(89))/2}}} or {{{x = (7-sqrt(89))/2}}}