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)

$5^{x+2}=25^{\frac{x+2}{2}}$
so:
$$25^{\frac{x+2}{2}}+25^{x+1}>750$$
$$25^{\frac{x}{2}}+25^x>30$$
Let's say $25^{\frac{x}{2}}=y$
$$y+y^2=30$$
completing the square:
$$(y+0.5)^2>30.25$$
$$y>\sqrt{30.25}-0.5=5$$
remeber that $25^{\frac{x}{2}}=y$
$$25^{\frac{x}{2}}>5$$
square both sides:
$$25^x>25$$
this means $x\geq1$
