4^x-(2*25^x)-10^x <0 solve the inequality?

Let u = 2^x and v = 5^x

4^x - (2 * 25^x) - 10^x <0
(2^2)^x - [2 * (5^2)^x] - (2 * 5)^x <0
(2^x)^2 - [2 * (5^x)^2] - (2^x * 5^x) <0
u^2 - 2 v^2 - u v < 0
u^2 - uv - 2 v^2 < 0
(u - 2v)(u + v) < 0
(2^x - 2*5^x)(2^x + 5^x) < 0

For all real values of x, 2^x + 5^x > 0
Then, 2^x - 2*5^x < 0
2^x < 2*5^x
log(2^x) < log(2*5^x)
x log(2) < log(2) + x log(5)
x log(2) - x log(5) < log(2)
x log(5) - x log(2) > -log(2)
x [log(5) - log(2)] > -log(2)
x > - log(2) / [log(5) - log(2)]

Consider the cases x = -1 and x = -(1/2).
At x = -1, you have 1/4 - 2/25 - 1/10 = 7/100.
At x = -1/2, you have 1/2 - 2/5 - 1/3.16...,
which is ABOUT 0.5 - 0.4 - 0.316, obviously < 0.
Therefore, there is some point between -1 and -1/2 where your expression would be exactly zero. Whether you get this by a series of successive guesses, or using Newton's method, I don't believe you can do it analytically; my answer is x = -0.7565....

When x > 0, the 2nd term is always the largest, so the expression is always negative.

When x << 0 (large and negative), the expression approaches 0, but can it change sign? As an example of what happens with a moderate-sized negative x, consider x = -3, which yields
1/64 - 2/15625 - 1/1000.
As you can see, all three terms will keep shrinking as x becomes more negative, but the positive term 4^x will shrink slower than the others, so the expression will remain positive as x --> -infinity.

The solution to the inequality is:
x > 0.7565...

4^x -2*25^x -10^x < 0 --------------(1)
solve 4^x - 2*25^x -10^x = 0
x= ln(2) / ln(2/5) =-0.76

http://www.wolframalpha.com/input/?i=solve+++4%5Ex+-+2*25%5Ex+-10%5Ex++%3D0

plot the point on the real line
--------- --------- -----------
-∞ .......-0.76........∞

Consider the intervals (-∞ ,-0.76),(-0.76, ∞ )
Choose any one point from each interval and test inequality (1)

(-∞ , -0.76) : choose x=-1
4^x - 2* 25^x -10^x < 0
7/100 < 0 (false)

(-0.76, ∞ ) : choose x=1
4^x - 2* 25^x -10^x < 0
-56 < 0 (true)

The solution is (-0.76 , ∞ )