How do you solve the quadratic $8b^2-1=-78$ using any method?

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Answer 1 · 2017-01-02T12:40:05

$b=(isqrt(77)sqrt(2))/4=(isqrt(154))/4 ~~i3.1024...$

Add 1 to both sides

$8b^2=-77$

Divide both sides by 8

$b^2=-77/8 -9.625 larr" as a check value"$

$b=sqrt(-77)/sqrt(2xx2^2)$

$b=(isqrt(77))/(2sqrt(2))xx1=(isqrt(77))/(2sqrt(2))xxsqrt(2)/sqrt(2)$

$b=(isqrt(77)sqrt(2))/4=(isqrt(154))/4 ~~i3.1024...$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check

$b^2=((isqrt(154))/4)^2$

$b^2=-154/16$

$b^2=-9.625 larr" check confirmed"$

How do you solve the quadratic 8b^2-1=-788b^2-1=-78 using any method?