Solve this exponential inequality $$
{5}^{(x+2)}+{25}^{(x+1)}>750\\
t=5^{(x+1)}\\
t^2+5t-750>0\\
t^2+5t-750=0\\
$$
$$
a=1, b=5, c= -750\\
D=35+3000=3025 \\
t_1= 25; t_2=-30
$$
$t_1=25\Longrightarrow x=1; t_2=-30$ Doesn't have a solution
So the solution is $x>1$.
Is this correct? I say is bigger than $1$ and not smaller than $1$ because $1$(numer before $t^2$ is bigger than $0$. Is this reason correct or is it because $5>1$ so the exponential function is progressive (the bigger the $x$, the bigger the y)

$$5^{(x+2)}  = 25 \cdot 5^x$$
$$25^{(x+1)} = 25 \cdot 25^x$$
$$750      = 25 \cdot 30$$
So your inequation may be expressed:
$$5^x + 25^x >30$$
As the derivative of $5^x + 25^x$ is always possitive, there is only one $x$ that makes the equality true:
$$x=1$$
And therefore, the solution of your inequation is $x>1$ 
