Answer:
We have, \[\sqrt{6x+7}-(2x-7)=0\] \[\sqrt{6x+7}=(2x-7)\] On squaring both sides \[{{\left( \sqrt{6x+7} \right)}^{2}}={{(2x-7)}^{2}}\] \[\Rightarrow 6x+7=4{{x}^{2}}+49-28x\] \[\Rightarrow 4{{x}^{2}}+42-34x=0\] \[\Rightarrow 2{{x}^{2}}-17x+21=0\] \[\Rightarrow 2{{x}^{2}}-14x-3x+21=0\] \[\Rightarrow 2x(x-7)-3(x-7)=0\] \[\Rightarrow (2x-3)(x-7)=0\] \[\Rightarrow x=\frac{3}{2}\] or \[7\] \[\therefore \,~x=7\] (as \[x=3/2\] doesn't satisfy the given equation)
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