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 $x=(-2sqrt6)/3$

How do you solve the quadratic 1-6x^2=-151-6x^2=-15 using any method?