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# x^2+6x-4=0. Solve with exact answers.?

Type exact answer, if you need to use radicals you can. Separate answers with commas.

### 6 Answers

- DaveLv 59 years agoFavorite Answer
x = (-b + sqrt(b^2 - 4ac)) / (2a)

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

x = (-6 + sqrt(36 - (-16))) / 2

x = (-6 + sqrt(36 + 16)) / 2

x = (-6 + sqrt(52)) / 2

x = (-6 + 2sqrt(13)) / 2

x = -3 + sqrt(13)

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

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

x = (-6 - sqrt(36 - (-16))) / 2

x = (-6 - sqrt(36 + 16)) / 2

x = (-6 - sqrt(52)) / 2

x = (-6 - 2sqrt(13)) / 2

x = -3 - sqrt(13)

So x equals:

-3 + sqrt(13)

and

-3 - sqrt(13)

- Anonymous9 years ago
easy peasy

- ✰✰Angel ✰✰Lv 49 years ago
x^2 + 6x - 4 = 0

Use quadratic formula

x= -b +_ square root of (b^2 - 4 ac)

-----------------------------------------------

2a

= - 6 +_ sqrt( (6)^2 - 4(1) (-4) )

------------------------------------------

2(1)

= -6 +_ sqrt(36+16)

--------------------------

2

= - 6 +_ sqrt(52)

--------------------

2

= -6 + 2 +_ (sqrt) 13 (The 52 is written as simpler. (2 root of 13)

------------------------------

2

= -3 +_ sqrt(13) -----> answer

Hope it helps!

Source(s): C$K - « Chippy »Lv 69 years ago
Quadratic formula: (-b±sqrt(b^2-4ac))/2a

a=1, b=6, c=-4

Therefore...

(-6±sqrt(6^2-4*1*(-4)))/(2*1) = (-6±sqrt(36+16))/2 = (-6±sqrt(52))/2

sqrt(52) can be written as 2*sqrt(13).

(-6±2*sqrt(13))/2=-3±sqrt(13)

And that's your answer. -3±sqrt(13)

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- Anonymous9 years ago
x²+6x-4 = 0

= (x²+6x+9)-13 = 0

= (x+3)²-(√13)² = 0

= (x+3+√13)(x+3-√13) = 0

As either of the multiplier must be 0, then x = -3+√13, -3-√13