How do you solve the quadratic using the quadratic formula given #10x^2-11x+9=14x-6x^2# over the set of complex numbers?

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Answer 1 · 2017-06-26T17:05:33

#x=15/32-isqrt351/32# or #15/32+isqrt351/32#

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

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

We have #10x^2-11x+9=14x-6x^2#

or #10x^2+6x^2-11x-14x+9=0#

or #16x^2-15x+9=0#

and using quadratic formula

#x=(-(-15)+-sqrt((-15)^2-4xx16xx9))/(2xx16)#

= #(15+-sqrt(225-576))/32#

= #(15+-sqrt(-351))/32#

= #15/32+-isqrt351/32#

i.e. #x=15/32-isqrt351/32# or #15/32+isqrt351/32#