Question 25461
It's not easy, but if I can do it, anyone can...sometimes I can't do it either.
take a good hard look at the first one.
There are lots of things to look for.
The highest power is 2. That means it's a quadratic equation. 
Maybe it's simple to factor, maybe not
this is the way I factor these things:
{{{2*k^2 + 3*k - 20}}}
what's the simplest factors of the first term?
They are 2k and k because {{{2*k^2 = (2*k) * k}}}
Now I set up the factors of the equation like this:
(k + _)(2k - _) 
Now fill in the blanks
what are the factors of the last term?
2 x 10 = 20
4 x 5 = 20
let's try 2 x 10
(k + 2)( 2k - 10)
each term in the first parenthesis gets multiplied by each term in the second
{{{2*k^2 + 4*k -10*k - 20 }}} =
{{{2*k^2 -6*k -20}}}
so that's not the answer
Let's swap the 2 and 10
(k + 10)(2k - 2)
{{{2*k^2 + 20*k -2*k - 20 }}} =
{{{2*k^2 +18*k -20}}}
that's not the answer either
Let's try 4 x 5
(k + 4)(2k - 5)
{{{2*k^2 + 8*k -5*k - 20 }}} =
{{{2*k^2 +3*k -20}}}
Hey, that's it!
Let's do the other one
{{{k^2 + 3*k - 4}}}
the factors of the first term are k and k because {{{k * k = k^2}}}
(k + _)(k - _)
fill in the blanks. What are factors of 4?
4 x 1 = 4
2 x 2 = 4
try 4 and 1
(k + 4)(k - 1)
{{{k^2 +4*k -k -4}}}
{{{k^2 +3*k -4}}}
that's it
now back to the original equation
2k2 + 3k - 20 divided by k2 + 3k - 4
substitute the factors for the equations
{{{((k + 4)*(2*k - 5))/((k + 4)*(k - 1))}}}
The (k + 4) factors are on the top and bottom, so they cancel
I end up with 
{{{(2*k - 5)/(k - 1)}}}
I think that's as simple as I can get it
Don't be afraid of this stuff. The more you do the easier it gets
It doesn't take a genius- just work.