Answer :
x, x+2 - two consecutive even integers
x is the smaller one
[tex]x^2=(x+2)+10 \\ x^2=x+12 \\ x^2-x-12=0 \\ x^2-4x+3x-12=0 \\ x(x-4)+3(x-4)=0 \\ (x+3)(x-4)=0 \\ x+3=0 \ \lor \ x-4=0 \\ x=-3 \ \lor \ x=4[/tex]
-3 isn't an even integer, so x=4.
[tex]x=4 \\ x+2=4+2=6[/tex]
The two integers are 4 and 6.
x is the smaller one
[tex]x^2=(x+2)+10 \\ x^2=x+12 \\ x^2-x-12=0 \\ x^2-4x+3x-12=0 \\ x(x-4)+3(x-4)=0 \\ (x+3)(x-4)=0 \\ x+3=0 \ \lor \ x-4=0 \\ x=-3 \ \lor \ x=4[/tex]
-3 isn't an even integer, so x=4.
[tex]x=4 \\ x+2=4+2=6[/tex]
The two integers are 4 and 6.
[tex](2n)^2=2n+2+10\\
4n^2=2n+12\\
4n^2-2n-12=0\\
2n^2-n-6=0\\
2n^2-4n+3n-6=0\\
2n(n-2)+3(n-2)=0\\
(2n+3)(n-2)=0\\
n=-\dfrac{3}{2} \vee n=2\\
-\dfrac{3}{2}\not \in \mathbb{Z}\Rightarrow n=2\\\\
2n=4\\
2n+2=6\\\\
\text{These numbers are 4 and 6.}
[/tex]
