Let the first number be x and y be the second.
Then we habe:
I) x - 4 = 2y
II) x*y = 30
We can rewrite equation I as:
x = 2y + 4
Applying this result on equation II, we have:
(2y + 4)*y = 30
2y² + 4y = 30
2y² + 4y - 30 = 0
y² + 2y - 15 = 0
![\begin{gathered} y_+=\frac{2+\sqrt[]{(2)^2-4\cdot1\cdot(-15)}}{2} \\ y_+=\frac{2+\sqrt[]{64}}{2} \\ y_+=(10)/(2) \\ y_+=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2qmyhholeeei8jliu5rs5sttyf2rswea13.png)
The second root is negative and don't satisfy equations I and II, then we can discard it.
Replacing y with 5 in equation I, we have:
x - 4 = 2*5
x = 2*5 + 4
x = 14