How do you solve #2p ^ { 2} - 22= 12p#?

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Answer 1 · 2018-04-20T11:06:32

#p=3+-2sqrt5#

#2p^2-22=12p#

#p^2-11=6p#

#p^2-6p-11=0#

Using the quadratic formula: #(-b+-sqrt(b^2-4ac))/(2a)#

#p=(6+-sqrt(6^2-4(1)(-11)))/(2(1))#

#p=(6+-sqrt(80))/2#

#p=(6+-4sqrt5)/2#

#p=3+-2sqrt5#