How do you solve #3x^2 + 5x + 1 = 0# using the quadratic formula?

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Answer 1 · 2018-08-04T06:46:53

#x = (-5 + sqrt13)/6# and #x = (-5 - sqrt13)/6#

#3x^2 + 5x + 1 = 0#

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

We know that:
#a = 3#, #b = 5#, and #c = 1# based on the equation, so let's plug them into the formula:

#x = (-5 +- sqrt(5^2 - 4(3)(1)))/(2(3))#

#x = (-5 +- sqrt(25 - 12))/6#

#x = (-5 +- sqrt13)/6#

Hope this helps!