Factor 18x^2+24x-10 please!!

When your front term has a coefficient you should look for a common factor, and then try factoring through grouping. In this case they all share a 2.

18x^2+24x-10
2(9x^2 + 12x - 5)

Now multiply the front and last terms
Product = 9(-5) = - 45

Next find the factors of this product that add up to the middle term = + 12
Factors of 45
1, 45
3, 15
5, 9

Since the product is negative the factors must have opposite signs, so 3 and 15 will give you 12
sum = -3 + 15 = 12

Now rewrite the quadratic with these two terms in the middle as coefficients of x, make two groups and factor

(2)(9x^2 -3x +15x - 5)
(2)((9x^2 -3x) +(15x - 5))
(2)(3x(3x - 1) + 5(3x - 1))
(2)(3x + 5)(3x - 1)

It seems like a lot of work but it gets much easier with time

The answer will look like this:

(ax + b)(cx - d)

If you expand this, you get
(a+c)x^2 + (bc - ad)x - bd

This forces you to use

a+c = 18
bc - ad = 24
bd = 10

Because you have only three equations for four unknowns, then you will have to use a bit of guesswork.

2(9x^2 + 12x - 5)
2(3x-1)(3x+5)

*** For the other answers, you can factor a 2 out of (6x-2)

(6x - 2)(3x + 5)

(3x + 5)(6x -2)