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---------(x^2+5)^2-15(x^¨2+5)+54=0-----------
Substitutions method
x^2+5=t
(t)^2-15(t)+54=0
Pq formula
t=-(-15/2)+-√((-15/2)^2-(+54))
t=7,5+-√(-56,25-54)
t=7,5+-10,5i
t1=18
t2=-3
>>>x^2+5=t
---------------Roots--------------------
x^2+5=18
x=3,6
x^2+5=-3
x=√-8
-----------------------------------------------
>>>Yet my textbook claims this is the answer
x=+-1
>>>What went wrong?
>>
bumb
>>
Nigger, if you're going to write more than two equations, you need to tex that shit.
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the fuck is tex nigger im from
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>>8582945
>the fuck is tex
Oh, you're an idiot. My bad.
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>>8582953
fucking nigger piss of my thread
>>
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>>8582956
fucking nigger piss of my thread
Wisdom
>>
Theres a minus sign in the root that shouldnt be there. Take it out and your t-roots will be 6 and 9,gives x-roots at +-2 and +-1.
Also 7.5+10.5i=18?
>>
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>>8582935
>t=-(-15/2)+-√((-15/2)^2-(+54))
>t=7,5+-√(-56,25-54)
>(-15/2)^2
>-56,25
A negative squared is a positive.
>>
do quadratic equation with t then put that value in what you subbed t for and you get x=2 and x=1
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