Our Quadratic is in the kind #ax^2+bx+c=0#. When we finish the square, us take half of the #b# value, square it, and add it to both sides.

First, let"s subtract #20# from both sides. We get:

#x^2+12x =-20#

Half that #12# is #6#, and when us square #6#, we gain #36#. Let"s add this to both sides. Us get:

#x^2+12x+36=-20+36#

Simplifying, we get:

#x^2+12x+36=16#

Now, we aspect the left side of the equation. Us think of 2 numbers that add up to #12# and have a product the #36#. #6# and #6# room our 2 numbers.

We can now variable this as:

#color(blue)((x+6)(x+6))=16#

Notice, what I have actually in blue is the very same as #x^2+12x+36#. We deserve to rewrite this as:

#(x+6)^2=16#

Taking the square root of both sides, us get:

#x+6=-4# and #x+6=4#

Subtracting #6# indigenous both sides of the equations, us get:

#x=-10, x=-2#


Answer connect
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Brandon
Mar 3, 2018

#x=-2# and #x=-10#


Explanation:

By completing the square, we always fifty percent the coefficient the #x#, putting it in brackets and squaring it, together a quadratic is always in the kind #ax^2+bx+c#. Always remember to include the consistent of #(+20)# top top the end:

#therefore# #x^2+12x -> (x+6)^2#

Also by perfect the square, we likewise take far the squared number of half of the ahead coefficient (the number in the brackets).

You are watching: Solve x2 + 12x + 6 = 0 using the completing-the-square method.

#(x+6)^2-36+20#

Simplifying terms:

#-36+20=-16#

Plugging this back in, removing the #-36# and #20#:

#(x+6)^2-16#

This is in the completed square form. You can always check this by broadening out:

#(x+6)(x+6) -> x^2+6x+6x+36-16 -> x^2+12x+20#

#therefore# This finish the square kind is correct.

Solving; so far we know:

#(x+6)^2-16=0#

So we add #16# to settle this, as we cannot solve when equal come 0.

#(x+6)^2=16#

As we execute not desire squares brackets, the contrary of squaring is square rooting, and this cancels out the squared brackets:

#(x+6)^2=16 -> x+6=pmsqrt16#

Always psychic the #pm# once square rooting...

As we desire #x# on that is own, we have to #-6# to eliminate it, delivering the #-6# to the other side the the equation.

#x+6=pmsqrt16 -> x=-6pmsqrt16#

Always remember there are mainly two solutions, as there is a plus and minus top top the answer.

#therefore# ours answers are...

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#x=-6pmsqrt16# and #x=-6pmsqrt16#

But utilizing our squared numbers:

#1^2=1xx1=1##2^2=2xx2=4##3^2=3xx3=9##4^2=4xx4=16# #sqrt16# can be simplified additional as it is a squared number...