How do you solve #x^2 + 5x + 6 = 0 # using the quadratic formula?

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Answer 1 · 2018-05-07T04:17:16

#x = -3#
#x = -2#

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

for a quadratic of the form #ax^2 + bx +c#.

We have #a = 1#, #b = 5#, and #c = 6#.

#x=(-5 +- sqrt(5^2-4(1)(6)))/(2(1))#

#x = (-5 +- sqrt(25-24))/(2)#

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

#x = (-5 +- 1)/(2)#

Hence,

#x = -6/2 = -3#
and
#x = -4/2 = -2#

I'm not sure why you wanted to use the quadratic formula, but you could just factor the quadratic:

#x^2+5x+6 = (x+3)(x+2) = 0#

#x+3 = 0 -> x = -3#
and
#x+2=0 ->x = -2#