SOLUTION: Please help me to solve this quadratic equation {{{3x - 2 ( x + 3 ) = 0}}}

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me to solve this quadratic equation {{{3x - 2 ( x + 3 ) = 0}}}      Log On


   

Question 785476: Please help me to solve this quadratic equation 3x+-+2+%28+x+%2B+3+%29+=+0
Answer by Trontor(4) About Me  (Show Source):
You can put this solution on YOUR website!
3x+-+2+%28+x+%2B+3+%29+=+0
Firstly, expand the 2%28x%2B3%29 using the distributive law, this makes our equation +3x+-+2x+%2B+6+=+0+
Since you didn't provide a method, I will do it the easiest way, the quadratic formula.
+3x+-+2x+=+1x+=+x
+x+%2B+6+=+0
From here you can say that x = -6 straight away, but using the QUADRATIC FORMULA:
The standard formula that applies to the quadratic formula is ax%5E2+%2B+bx+%2B+c+=+0+
a = Coefficient of x%5E2 b = Coefficient of x c = the single number value
a = 0 (1x%5E2+=+0) b = 1 (x+=+1x) c = 6
Quadratic Formula = x+=+%28-b+%2B-+sqrt%28+b%5E2-%284%2Aa%2Ac%29+%29%29%2F%282%2Aa%29+
WE CANNOT DO THIS! AS 2*a = 0 , which means we CANNOT divide by zero.
I will assume that your equation is %283x+-+2%29+%28+x+%2B+3+%29
Which is expanded to 3x%5E2+%2B+7x+-+6+=+0 which is fine for a quadratic equation.
a = 3 b = 7 c = -6
Now we substitute the a, b, and c values that are in your equation into this quadratic formula so it can return 2, (maybe 1 in this case) values of x.
x+=+%28-7+%2B-+sqrt%28+7%5E2-%284%2A3%2A-6%29%29%29%2F%282%2A3%29+
x+=+%28-7+%2B-+sqrt%2849+-+-72%29%29+%2F+6
x+=+%28-7+%2B-+sqrt%2849+%2B+72%29%29+%2F+6
x+=+%28-7+%2B-+sqrt%28121%29%29+%2F+6
x=+%28-7+%2B-+11%29+%2F+6
THEREFORE
x+=+%28-7+%2B+11%29+%2F+6
x++=+4%2F+6
x+=+0.67%7D%7D%0D%0Aor+%0D%0A%7B%7B%7Bx+=+%28-7+-+11%29+%2F+6
x+=+-18%2F+6
x+=+-3%7D%7D%0D%0A%0D%0A%7B%7B%7Bx+=+0.67+or+-3
I personally think you mistyped the equation, it was %283x+-+2%29%28+x+%2B+3+%29+=+0
not 3x+-+2+%28+x+%2B+3+%29+=+0.
Thanks, remember to mark this as your solution if I got it!