How do you solve #sqrt(5x+39) = x+3#?

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Answer 1 · 2016-08-13T15:34:52

#x={-6,5}#

#sqrt(5x+39)=x+3#

#(sqrt(5x+39))^2=(x+3)^2#

#5x+39=x^2+6x+9#

#x^2+6x-5x+9-39=0#

#x^2+x-30=0#

#"Factor the equation"#

#(x-5)(x+6)=0#

#"if "(x-5)=0" ; "rarr x=5#

#"if "(x+6)=0" ; "rarr x=-6#

#x={-6,5}#