How do you solve #p^ { 2} - 35= 7+ p#?

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Answer 1 · 2017-08-05T06:56:00

#p=-6 or p=7#

#p^2-35=7+p#
#=>p^2-p-42=0#
#=>(p+6)(p-7)=0#
#=>p=-6 or p=7#

Answer 2 · 2017-08-05T06:58:20

#p=7 or p=-6#

Bring all terma on one side, #p^2-p-35-7=0#

#p^2-p-42=0#

The quadratic formula is #p=(-b+-sqrt(b^2-4ac))/(2a)#

Here we have: #p=(-(-1)+-sqrt((-1)^2-4(1*-42)))/(2*1)#

#p=(1+-sqrt(1-4(-42)))/2#

#p=(1+-sqrt(1+168))/2#

#p=(1+-13)/2#

#p=14/2=7 or p=-12/2=-6#