3x^2+6x+2=0 could someone help me with this, solve using quadratic formula.?

Answer 1

If the quadratic is of the form
ax^2 + bx + c
The solution with the quadratic formula is

x= [-b +/- sqrt(b^2 - 4ac)]/2

In this case a=3, b=6, c=2, so

x = [-6 +/- sqrt(6^2-4*3*2)]/2

so
x=-1.268 or x=-4.732

Answer 2

standard quadratic form:

a x^2 + b x + c = 0

quadratic formula:

x = [ -b +/- SQRT(b^2 - 4 ac)] / 2a


Where +/- means "plus or minus" as there can be two solutions.
SQRT means square root.
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In your problem, a = 3, b = 6, c = 2

x = [ -6 +/- SQRT(36 - 4*3*2)] / 6
x = [ -6 +/- SQRT(12)]/6
x = -6/6 +/- SQRT(12) / 6

SQRT(12) = SQRT(4*3) = SQRT(4)*SQRT(3) = 2*SQRT(3)
SQRT(12) / 6 = (1/3)*SQRT(3) = approx. 0.57735...

x = -1 + (1/3)SQRT(3) and
x = -1 - (1/3)SQRT(3)

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One shortcut I use is to make a = 1/2 (then 2 a = 1)

To make a = 1/2, I would divide your equation by 6 thoughout. This is no problem for the right-hand side, as it is zero.

(1/2)x^2 + x + (1/3) = 0

x = -1 +/- SQRT[1^2 - 4(1/2)(1/3)]
(I don't need to divide by 2a since that is 1)

x = -1 +/- SQRT(1 - 2/3)
x = -1 +/- SQRT(1/3)
x = -1 +/- 1/√3

But we are often told not to leave a root in the denominator, so we multiply the second part by √3/√3 (this is the same as multiplying by 1, so no change).

1/√3 } √3/√3 = √3 / 3

x = -1 +/- (1/3)√3
(which is the same as what we had above, a relief).

Answer 3

The answer is

x = [- 6 ± Sqrt (6 - 4*3*2) ] / (2*3)

using the formula which is

x = [- b ± Sqrt (b - 4*a*c) ] / (2*a)

and a = 3, b = 6, and c = 2 taken from your above equation.

Answer 4

the formula is x = -b+sqrt(b^2-4ac) / 2a

a = 3, b = 6 and c = 2

b^2 - 4ac = 36 - 24 = 12

x = -6 + 2sqrt(3) / 6 = -1 + sqrt(3)/3
or -6 - 2sqrt(3) / 6 = -1 - sqrt(3)/3

use your calc if you want the ans in decimals.

Answer 5

Quadratic formula is x= (-b+-(b^2-4ac)^(1/2))/2a, so:
(-6+-(36-4(3)(2))^(1/2))/(2(3))
(-6+-(12)^(1/2))/6
(-6+-2rad3)/6, which can be simplified down to
(-3+-rad3)/3

Answer 6

3x^2+6x+2=0


D= b^2 - 4ac = 36 - 24 = 12


-b +/- SQRT(D) -6 +/- 2 SQRT(3)
----------------- = ---------------- = (6, -1)
2a 6


so the solutions are

-1 - [2 SQRT(3)]/6

-1 + [2 SQRT(3)]/6

or

(x+1 + [2 SQRT(3)]/6) and (x + - [2 SQRT(3)]/6)

Answer 7

delta = 6^2 - 4*2*3 = 36 - 24 = 12
x1 = [-6 +sqrt(12)]/6 = -1 + sqrt(3)/3
x2 = [-6 - sqrt(12)]/6 = -1 - sqrt(3)/3

Answer 8

x={-6+-sq rt(6^2-4.3.2)}/2.3
or x={-6+-sq rt(36-24)}/6
or x=-1+-sq rt(12)/6
or x=-1+-sq rt(3.2^2)/6
or x=-1+-2sq rt(3)/6
or x=-1+-sq rt(3)/3 ans