How do you find the solution to the quadratic equation # 0 = x^2 + 5x + 6#?

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Answer 1 · 2018-06-20T17:34:21

x=-2 or -3

#x^2+5x+6=0#

#x^2+2x+3x+6=0#

#x*(x+2)+3(x+2)=0#

#(x+2)*(x+3)=0#

#x=-2 or -3#

Answer 2 · 2018-06-20T18:13:05

#x=-3, x=-2#

Lets factorise:

#(x+3)(x+2)=0#

Check this is correct...

#x^2+3x+2x+6#

#rArr x^2+5x+6=0#

Now solve:

#(x+3)=0#

#therefore# #x=-3#

#(x+2)=0#

#therefore# #x=-2#

Answer 3 · 2018-06-21T23:06:21

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

Are there any two numbers that sum to the middle term (#5#), and have a product of the last term (#6#)?

After some trial and error, we arrive at

#2# and #3#. Thus, we can factor the right side of our quadratic as

#0=(x+2)(x+3)#

Setting both of our factors equal to zero, we get

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

Hope this helps!