How do you solve the quadratic #1-6x^2=-15# using any method?

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Answer 1 · 2016-09-03T03:19:28

#x=(2sqrt6)/3# or #x=(-2sqrt6)/3#

#1-6x^2=-15#

#hArr6x^2-15-1=0# or

#6x^2-16=0#

According to quadratic formula solution of equation #ax^2+bx×c=0# is given by

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

Hence, solution of #6x^2-16=0# is

#x=(-0+-sqrt(0^2-4×6×(-16)))/(2×6)#

= #+-sqrt(4×6×16)/12#

= #+-8sqrt6/12#

= #+-(2sqrt6)/3#

Hence solution of equation is #x=(2sqrt6)/3# or #x=(-2sqrt6)/3#