Answer:
We have, \[{{x}^{2}}-2ax-(4{{b}^{2}}-{{a}^{2}})=0\] \[{{x}^{2}}-2ax+{{a}^{2}}-4{{b}^{2}}=0\] \[{{(x-a)}^{2}}-{{(2b)}^{2}}=0\] \[\therefore \,\,(x-a+2b)(x-a-2b)=0\] \[\Rightarrow x=a-2b\] or \[a+2b\] Hence, \[x=a-2b\] or \[x=a+2b\]
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