hi could anybody show me how to factorise this quadratic please.
3x^+5x^+2=0
thanks for any help.

I guess you mean 3x^2 + 5x - 2=0?

I'll go through it the method I use.

1. First, look for the factors of the coefficient of x^2 and the constant.
3x^2 = 3x \times x or -3x \times -x
-2 = -2 \times 1 or 2 \times -1

2. I'll take the first possibilities for 3x^2 and -2.
3x\,\,\,\,\,x \\ -2\,\,\,\,\,1

3a. Multiply one term from the upper ones by one of the lower ones, then add both results:
3x\,\,\,\,\,\,\,\,\,x \\ -2\,\,\,\,\,\,\,\,\,1 \\ \\ -6x\,\,\,\,\,x= -5x

3b. As you can see, the final result is -5x, which is not the 5x that is present in the original equation. Ok, multiply the terms diagonally:
3x\,\,\,\,\,\,\,\,\,x \\ -2\,\,\,\,\,\,\,\,\,1 \\ \\ -2x\,\,\,\,\,3x= x

4a. As you can see, the final result is -5x, which is not the 5x that is present in the original equation. Ok, change the factors (from the ones in 1.) . Use the first and second ones.
3x\,\,\,\,\,x \\ 2\,\,\,\,\,-1

4b. Do the same thing as in step 3a.
3x\,\,\,\,\,\,\,\,\,x \\ 2\,\,\,\,\,\,\,\,\,-1 \\ \\ 6x\,\,\,\,\,-x= 5x

Here! You got the 5x that you had in the original equation. So, Bracket the diagonal terms together.
(3x-1)(x+2)

If you expand, you'll have the original expression back.

Now, rewrite everything:
3x^2 + 5x - 2 = (3x-1)(x+2) = 0

Since any number multiplied by zero gives zero, either (3x-1) is zero, or (x+2) is zero. You have two equations then:
3x-1 = 0 or x+2 =0

Solve for both. Tell us what you get.

I hope it helped! :)