Answer:
form of devision that takes longer bc it a fraction or decimal. first you want to get rid of the decimals. then devide the numbers from the lowest to the highest
I may be wrong been for ever since I did it on paper
The picture is omitting the question

$\implies 2^{2x}-2\cdot5^{2x}-(2^x)(5^x)>0$
let $2^x=a$ and $5^x=b$
So, $a^2-2b^2-ab>0$
divide by $b^2$, ($b^2>0$)
$\implies \left(\frac ab\right)^2-\left(\frac ab\right) -2>0$
this is a quadratic, in $\left(\frac ab\right)$, let it be $x$
So, $x^2-x-2>0$
Can you simplify it now?