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sqrt{9x+67}=x+5
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Is this the question? \[\sqrt{9x+67}=x+5\]
Square both sides of the equation \[(\sqrt{9x+67})^2=(x+5)^2\]
Square both sides & solve it:)
The squares cancel \[9x+67=x^2+10x+25\] Move all terms to one side and cancel \[0=x^2+10x-9x+25-67\]
Are you able to take it from there @pinkwhispers80?
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\[0 =x^2+x-42 \] \[(x+7)(x-6)=0\] According to the zero factor theorem either x=6 or x=-7
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