solve the following equation: x^2+6x+3=0?

Question

i just can't figure this one out i need help can anyone help to understand?

Answer 1

Solve using Sreedharacharya's formula for solving quadratic equations.
According to the formula, the roots of the quadratic equation ax^2 + bx + c = 0 are
x = {-b + sqrt(b^2 - 4ac)}/(2a), {-b - sqrt(b^2 - 4ac)}/(2a)

Here, a = 1, b = 6, c = 3.

Hence, x = {-6 +/- sqrt(6^2 - 4*1*3)} / 2
x = {-6 +/- sqrt(24)} / 2
x = -3 + sqrt(6), -3 - sqrt(6)

Answer 2

Completing the square

x² + 6x + 3 = 0

Transpose 3

x² + 6x + 3 - 3 = 0 - 3

Collect like terms

x² + 6x = - 3

x² + 6x +_____= - 3 +_____

x² + 6x + 9 = - 3 + 9

Collect like term

x² + 3x + 9 = 6

(x + 3)(x + 3) = 6

)x + 3)² = 6

(√x + 3)² = ± √6

x + 3 = ± √6

Transpose 3

x + 3 - 3 = - 3 ± √6

Collect like terms

x = - 3 ± √6

x = - 3 ± 2.449489743

- - - - - - - - - -

Solving for +

x = - 3 + 2.449489743

x = 0.550510257

- - - - - - - -

Solving for -

x - 3 - 2.449489743

x = 5.449489743

- - - - - - -

You can also solve this equation by using the quadratic formula.

- - - - - - -s-

Answer 3

OK

Doesn't look like you can factor so ...
quadractic formula

-6 +-sqrt(36 - 4(1)(3)) / 2

-6 +- sqrt(36 -12) /2
-6 +- sqrt(24) /2
-6 +- sqrt(4*6) /2
-6 +-2sqrt(6)/2

-3 +-sqrt6
so you have two roots

-3 + sqrt6 and -3 -sqrt6

Proof

(-3+sqrt6)^2 + 6(-3 +sqrt6) + 3 = 0??
9 -6sqrt6 +6 -18 +6sqrt6 +3 = 0??
9 +6 -18 +3 = 0??
18-18 = 0??
0 = 0 YES!!
Hope that helps.

Answer 4

You must be doing the quadratic formula since this doesn't factor.
a = 1
b = 6
c = 3

b^2 - 4ac = 36 - 4*3*1 = 24

x = (-6 +/- sqrt(24))/2
or as decimals
x = -.55 or x = -5.45

Answer 5

use quadratic formula, youll get the 2 answers to be -3 + sqrt(6) and -3 - sqrt(6)