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\sqrt{6n+19}=2-\left(-n\right)
Subtract -n from both sides of the equation.
\sqrt{6n+19}=2+n
Multiply -1 and -1 to get 1.
\left(\sqrt{6n+19}\right)^{2}=\left(2+n\right)^{2}
Square both sides of the equation.
6n+19=\left(2+n\right)^{2}
Calculate \sqrt{6n+19} to the power of 2 and get 6n+19.
6n+19=4+4n+n^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+n\right)^{2}.
6n+19-4=4n+n^{2}
Subtract 4 from both sides.
6n+15=4n+n^{2}
Subtract 4 from 19 to get 15.
6n+15-4n=n^{2}
Subtract 4n from both sides.
2n+15=n^{2}
Combine 6n and -4n to get 2n.
2n+15-n^{2}=0
Subtract n^{2} from both sides.
-n^{2}+2n+15=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=2 ab=-15=-15
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn+15. To find a and b, set up a system to be solved.
-1,15 -3,5
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.
-1+15=14 -3+5=2
Calculate the sum for each pair.
a=5 b=-3
The solution is the pair that gives sum 2.
\left(-n^{2}+5n\right)+\left(-3n+15\right)
Rewrite -n^{2}+2n+15 as \left(-n^{2}+5n\right)+\left(-3n+15\right).
-n\left(n-5\right)-3\left(n-5\right)
Factor out -n in the first and -3 in the second group.
\left(n-5\right)\left(-n-3\right)
Factor out common term n-5 by using distributive property.
n=5 n=-3
To find equation solutions, solve n-5=0 and -n-3=0.
-5+\sqrt{6\times 5+19}=2
Substitute 5 for n in the equation -n+\sqrt{6n+19}=2.
2=2
Simplify. The value n=5 satisfies the equation.
-\left(-3\right)+\sqrt{6\left(-3\right)+19}=2
Substitute -3 for n in the equation -n+\sqrt{6n+19}=2.
4=2
Simplify. The value n=-3 does not satisfy the equation.
n=5
Equation \sqrt{6n+19}=n+2 has a unique solution.