Help.

You read it wrong!!! Oops!

It said difference, not sum

This is my way, there might be a better way.

When it says "their product is 45" I just find its factor pairs

5 * 9

15 * 3

Those are the first to come into my mind

Then I instantly recognize 15 and 3 has a DIFFERENCE of 12.

You can make a system of equations, but that is too slow.

the difference must be 15 - 3 = 12. All steps you do are correct except the last one (18 = 15 + 3)
^{_{friv 2020}}

Help.

$\begin{array}{|lrcll|} \hline (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline (1)-(2): & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- (a^2-2ab+b^2) \\ & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- a^2+2ab-b^2 \\ & (a+b)^2-(a-b)^2 &=& 4ab \\ & (a+b)^2 &=& 4ab+(a-b)^2 \\ & \mathbf{ a+b } &=& \mathbf{\sqrt{4ab+(a-b)^2}} \quad | \quad ab=45,\quad a-b = 12 \\ & a+b &=& \sqrt{4\cdot 45+12^2} \\ & a+b &=& \sqrt{324} \\ &\mathbf{ a+b} &=& \mathbf{18} \\ \hline \end{array}$

The sum of the integers is 18

Let a be the larger integer and b be the smaller....so...we have

a - b  =  12   and   ab  = 45

Rewrite the first equation  as  a = 12 + b

Sub this into the second for a

(12 + b) b  = 45      simplify

12b + b^2 =  45     rearrange as

b^2 + 12b  - 45  = 0

We need two factors that multiply to -45   and sum to 12

These are  15  and -3

So we write

(b +15) (b -3)  = 0  set each factor to 0  and solve for b and we get that b = -15  (reject)  and b = 3

And .... a   =  12 + b  =   12 + 3   = 15

And their  sum  is  18