How do you solve this eqution ? 3-\sqrt {3n^{2}+n-3}=n+1

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Answer 1 · 2017-06-03T19:32:23

Assumption: The question should be:
$" " 3-sqrt(3n^2+n-3)=n+1$

Taken to a point where you should be able to finish it off.

$3-n-1=sqrt(3n^2+n-3)$

Square both sides

$(2-n)^2=3n^2+n-3$

$4-4n+n^2=3n^2+n-3$

$2n^2+5n-7=0$

Using standard form $y=an^2+bn+c$

Where $" "a=2"; "b=5"; "c=-7$

and $n= (-b+-sqrt(b^2-4ac))/(2a)$

$=>n" "=" "(-5+-sqrt(5^2-4(2)(-7)))/(2(2))$

$n" "=" "-5/4+-sqrt(25+56)/4$

I will let you finish this off.