value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$ 
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$

$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-3}{5^x-1}\right) = -\frac{(5^x+3)(5^x-1)}{5^x-1} = -(5^x+3)$$
Now how can i solve after that help required, Thanks

$z=5^x \to $ $z^2+(a+2)z-(a+3) \lt 0$
now $ \\ \Delta = \to (a+2)^2-4(1)(-(a+3)) = \\a^2+4+4a+4a+12=a^2+8a+16=(a+4)^2 \geq0 \\$
now $$z=\frac{-(a+2)\pm \sqrt{(a+4)^2}}{2}=\\z=1 ,z=-(a+3)\\z=1 \to 5^x=1 \to x=0\\ (5^x-1)(5^x+a+3)<0$$
