Question

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

Answer 1

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]