solve for x x^2+12x+5=0?

Answer 1

x^2+12x+5=0
By using quadratic formula
a=1 , b=12 ,c= 5
x=-12+/- sq rt{144-20}
x=[-12+/- sq rt{124}]/2
x=[-6+/- sq rt{31}]

Answer 2

10

Answer 3

The answers, I'm fairly sure, are -6 + sqrt(31) and -6 - sqrt(31).

The only way to solve this is through use of the quadratic formula:

[-b (+ or -) sqrt(b^2 - 4ac)] / 2a

Where a is the coefficient of the squared term, b is the coefficient of the x term, and c is the constant.

Answer 4

x^2+12x+5=0

x^2+12x=-5

x^2 + 2.x.6 + 6^2 = -5 + 6^2

(x+6)^2 = -5 + 36

(x+6)^2 = 31

x+6 = square root of 31

x= square root of 31-6

or

method 2.

let a=1, b=12, c=5

then x= [ -b (+) or(-) {b^2 -4ac}^1/2]/2a

by substituting, u get the value of 'x'

Answer 5

foil it out. what multiple of (1) from x^2 and (5) constant add up to equal 12.
Apparently none that is noticable.

Use the quadratic formula then. I dont have a calc :(

Small hints tho, it has to be negative and between -12 and 0.

Answer 6

2x^2+12x+5=0

Just a shot n the dark, been out of school for years.

Answer 7

x^2 + 12x + 5 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-12 ± sqrt(12^2 - 4(1)(5)))/(2(1))
x = (-12 ± sqrt(144 - 20))/2
x = (-12 ± sqrt(124))/2
x = (-12 ± sqrt(4 * 31))/2
x = (-12 ± 2sqrt(31))/2

x = -6 + sqrt(31) or -6 - sqrt(31)

Answer 8

compare equation to general eq ax^2+bx+c=0

So b^-4ac=144-20=124

So one sol is = (-b+sqrt(b^2-4ac))/2a = (-12+sqrt(124))/2 = (-6+sqrt(31)
Other sol is = (-b-sqrt(b^2-4ac))/2a = (-12-sqrt(124))/2 = (-6-sqrt(31)

So Sols are:
(-6+sqrt(31)) and (-6-sqrt(31))

Answer 9

minus b plus or minus square root of b squared minus 4ac all over 2a gives you -6 +or- (SQR124)/2.

Answer 10

quadratic formula will solve de problem