How do you solve #4x^2+12x+3=0#?

1 Answer
George C.
Aug 7, 2016

Use the quadratic formula to find:

#x = (-3+-sqrt(6))/2#

Explanation:

#4x^2+12x+3=0# is in the form #ax^2+bx+c=0# with #a=4#, #b=12# and #c=3#

We can solve it using the quadratic formula:

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

#= (-12+-sqrt(12^2-4(4)(3)))/(2*4)#

#= (-12+-sqrt(144-48))/8#

#= (-12+-sqrt(96))/8#

#= (-12+-sqrt(4^2*6))/8#

#= (-12+-4sqrt(6))/8#

#= (-3+-sqrt(6))/2#

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