Solve using the quadratic formula

[MickieMik]

x^2 - 3x = - 6x - 1

Then I did this
x^2 - 3x + 6x +1 to get
x^2 +3x +1= 0
a=1. b = 3, c = 1
so I did this
x = -3 +/- sq rt [3^2 - 4(1)(1)] / 2(1) to get
x = -3 +/- sq rt [9 - 4] / 2 then
x = -3 +/- sq rt [5] /2
you can't square it so is this the answer?

Help??

 

[Violagirl]

Yep, you got the right answer. When you have a number that doesn't have an even square root, you can leave it as your final answer.

 

[Deleted member 4993]

x^2 - 3x = - 6x - 1

Then I did this
x^2 - 3x + 6x +1 to get
x^2 +3x +1= 0
a=1. b = 3, c = 1
so I did this
x = -3 +/- sq rt [3^2 - 4(1)(1)] / 2(1) to get
x = -3 +/- sq rt [9 - 4] / 2 then
x = (-3 +/- sq rt [5] )/2 <<<< without the parentheses - your answer will be counted wrong.
you can't square it so is this the answer?

Help??