How do you solve #x^2-5x-8=0#?

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How do you solve #x^2-5x-8=0#?

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Answer 1 · 2016-05-10T10:45:49

The roots are #6.275, -1.275#

Explanation:

This is a quadratic equation of form #ax^2+bx+c# where #a=1;b=-5;c=-8 ; b^2-4ac=25+32=57(+)#ive. So roots are real. #x= -b/(2a) +- sqrt (b^2-4ac)/(2a)= 2.5 +- sqrt57/2 = 6.275, -1.275# [Ans]

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How do you solve #x^2-5x-8=0#?

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