How do you solve the quadratic using the quadratic formula given #10x^2+9=x#?

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Answer 1 · 2016-08-31T15:17:21

#x=1/20-isqrt359/20# or #1/20+isqrt359/20#

Quadratic formula gives the solution of quadratic equation #ax^2+bx+c=0# as

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

Hence solution of the equation #10x^2+9=x# which can also be written in general form as #10x^2-x+9=0# is

#x=(-(-1)+-sqrt((-1)^2-4×10×9))/(2×10)#

= #(1+-sqrt(1-360))/20#

= #1/20+-isqrt359/20#

i.e. #x=1/20-isqrt359/20# or

#1/20+isqrt359/20#