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\[\sqrt{x+14}-12=x\] \[\sqrt{x+14}=x+12\] square both sides get \[x+4=x^2+24x+144\] solve the quadratic \[x^2+20x+140=0\] and check the solutions
oops no no solve the quadratice \[x^2+23x+140=0\]
we can test it \[\sqrt{-10+14}-12=-10\] \[\sqrt{4}-12=-10\] \[2-12=-10\] yup
yw
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