Simple problem

suppose, the largest number=S and the other number is=Y

The difference between two numbers is 2, so, $X-Y=2$ ...........................[1]

Their product is 84 greater than the square of the smaller number, so, $XY=Y^2+84$ or, $Y(X-Y=84$ ....................[2]

now, dividing equation [2] by equation [1] , we get,

$\frac{Y(X-Y)}{X-Y}$ = $\frac{84}{2}$

or, $Y=42$ , so, $X=44$

now, $X+Y=42+44=86$

Let $x$ be the larger number and $y$ be the smaller number. Then we have

$x-y=2$ $\implies$ $x=y+2$ $\color{#D61F06}(1)$

$xy=84+y^2$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ . We have

$(y+2)y=84+y^2$

$y^2+2y=84+y^2$

$2y=84$

$y=42$

It follows that $x=42+2=44$ . So the desired sum is $42+44=\boxed{86}$

For sake of variety, and to find $x + y$ directly without first finding the values for $x,y$ ....

With $x \lt y$ we are given that $y - x = 2$ and $xy = x^{2} + 84.$ Now

$(y - x)^{2} = 4 \Longrightarrow y^{2} + x^{2} - 2xy = y^{2} + x^{2} - 2(x^{2} + 84) = 4 \Longrightarrow y^{2} - x^{2} = 4 + 2*84 = 172$

$\Longrightarrow (y - x)(y + x) = 172 \Longrightarrow y + x = \dfrac{172}{y - x} = \dfrac{172}{2} = \boxed{86}.$

Let smaller no. be \color{ green}{y} and larger be $\color{#D61F06}{x}$
According To Question
● $\color{#20A900}{y}-\color{#D61F06}{x}=2....(1)$
● $\color{#D61F06}{x}×\color{#20A900}{y}=\color{#D61F06}{x^2}+84....(2)$
● $\color{#D61F06}{x}×(\color{#D61F06}{x}+2)=\color{#D61F06}{x^2}+84$
● $\color{#D61F06}{x^2+2}\color{#D61F06}{x}=\color{#D61F06}{x^2}+84$
● $\color{#D61F06}{x}=\boxed{42}$ $and$ $\color{#20A900}{y}-42=2=\boxed{44}$
● $Now,\color{#D61F06}{x}+\color{#20A900}{y} =42+44=\boxed{86}$