Let's call the first number N1 and the second number N2.
So, let's call N1 the smaller number and N1=x
Let's call N2 the larger number and N2=x+4.
X(X+4)=45
x^2+4x=45
x^2+4x-45=0
(x+9)(x-5)=0
X=-9, X=5
We will assume that the numbers are x and y.
Given that the difference between the numbers is 4.
==> x - y = 4 .............(1)
Also, given the product of the numbers is 45.
==> x*y = 45................(2)
Now we will use the substitution method to solve.
From (1), we know that x = y + 4.
==> x*y = 45
==> (y+4)*y = 45
==> y^2 + 4y = 45
==> y^2 + 4y - 45 = 0
Now we will solve the quadratic equation using factors.
==> (y+ 9)(y-5) = 0
==> y1= -9 ==> x1= -5
==> y2 = 5 ==> x2= 9
Then the numbers are 5 and 9 OR -5 and -9.
The two numbers to be determined have a difference of 4, so let them be A and A + 4.
Now we are given that their product is 45
=> A*( A + 4) = 45
=> A^2 + 4A = 45
=> A^2 + 4A - 45 = 0
=> A^2 + 9A - 5A - 45 = 0
=> A( A + 9) - 5(A + 9) = 0
=> (A - 5)(A +9) = 0
This gives A = 5 and A = -9
So the two numbers are 5 , 9 and -9, -5
The required numbers can be (5, 9) and (-9, -5)