If  $5^{2+4+6+\cdots +2x}={\left(0.04\right)}^{-28}$ and  $(x>0)$,  then  $x=$

$5^{2+4+6+\cdots +2x}={\left(0.04\right)}^{-28}$

$5^{2(1+2+3+\cdots +x)}={\left(\frac{1}{25}\right)}^{-28}$

$5^{2\times \frac{x(x+1)}{2}}={\left(\frac{1}{25}\right)}^{-28}={\left(5^{-2}\right)}^{-28}$

$5^{x(x+1)}=5^{56}$

$x(x+1)=56$

$x^2+x-56=0$

$x^2+8x-7x-56=0$

$(x+8)(x-7)=0$

X= -8 (not possible), x=7 possible