How do you solve #-2p ^ { 2} = 12p + 15#?

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Answer 1 · 2017-05-19T20:34:46

#x=(-6+sqrt(6))/2#

Move #-2p^2# to the other side:

#0=2p^2+12p+15#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug into this formula:

#x=(-12+-sqrt(12^2-4(2)(15)))/(2(2))#

Simplify:

#x=(-12+-sqrt(144-120))/4#

#x=(-12+-sqrt(24))/4#

#x=(-12+-2sqrt(6))/4#

#x=(-6+sqrt(6))/2#