How do you solve #12p^2-12 p-13=0#?

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Answer 1 · 2018-06-09T15:52:27

#x=(3+-4sqrt3)/6#

Given: #12p^2-12p-13=0#.

Use the quadratic formula, which states that:

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

Here, #a=12,b=-12,c=-13#.

So we get:

#x=(12+-sqrt(144-4*12*-13))/24#

#=(12+-sqrt(144+624))/24#

#=(12+-sqrt768)/24#

#=(3+-4sqrt3)/6#

Answer 2 · 2018-06-09T15:57:57

#x=(3+-4sqrt(3))/(6)#

You will need to use the Quadratic Formula to solve this quadratic in the form:

#ax^2+bx+c#

#12p^2-12p-13=0#

#a=12#

#b=-12#

#c=-13#

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

#x=(-(-12)+-sqrt((-12)^2-4*12*(-13)))/(2*12)#

#x=(12+-sqrt(144-(-624)))/(24)#

#x=(12+-sqrt(768))/(24)#

#x=(12+-16sqrt(3))/(24)#

#x=(4(3+-4sqrt(3)))/(24)#

#x=(3+-4sqrt(3))/(6)#