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)

Go slower! First note that $25^x=(5^x)^2$ and set $t=5^x$. Then substitute:
$$
5^{x+2}=5^x\cdot 25=25t,
\qquad
25^{x+1}=25^x\cdot25=25t^2
$$
Thus the inequality becomes
$$
25t+25t^2>750
$$
and, simplifying,
$$
t^2+t-30>0
$$
or
$$
(t+6)(t-5)>0
$$
