Please help me find two numbers

Find two numbers whose sum equals 38 and whose product equals 358 .thanks

lets express the 2 numbers as variables x and y.

two numbers whose sum is 38: x+y=38

whose product (multiplication) is 358: x*y=358

x+y=38
x*y=358

solve for x in the first equation then plug the value of x into the second equation
x+y=38 <=== subtract both sides by y
x+y-y=38-y
x=38-y

replace the x in the second equation with 38-y
x*y=358
(38-y)y=358 <=== distribute the y
38y-y^2=358 <=== bring 358 to the other side by subtracting and rearrange the equation
-y^2+38y-358=0
using quadratic formula you should get
y=19+sqrt(3)
y=19-sqrt(3)

plug both those in x=38-y
x=38-(19+sqrt(3))
x=38-19-sqrt(3)
x=19-sqrt(3)

x=38-(19-sqrt(3))
x=38-19+sqrt(3)
x=19+sqrt(3)

The two numbers are:
19+sqrt(3) and 19-sqrt(3)

The question is not going to work with any type of number, in fact I believe that the answer is an IMAGINARY, nail that in your head son, put both formats in algebraic form so

x + y = 38

xy = 358

substitute for each integer.

so y = (358) / (x)

and then substitute it, so

x + (358/x) equals 38

so x^2 plus 358 over x equals 38

times x on each side and you get x^2 plus 358 equaling 38x

minus the 38x on each side and you get a quadratic equation. x^2 - 38x + 358

Now use the quadratic formula

-b plusminus sqrt b^2 - 4ac / 2a

a equals 1, b equals -38, and c equals 358

now solve it for yourself or stay dumb either way.

-_-

let numbers are x & y
x+y = 38 , xy = 358 ====> y=358/x , put in first equation
358/x + x = 38 ===> x^2 -38 x +358 =0
x = [38 ± {(38)^2 - 4*358}^1/2] / 2...............by quadratic formula
=[38 ±sqrt(12)] / 2 = 19 ± sqrt(3)
so corresponding to this x ,
y= 19 - sqrt(3) , 19 +sqrt(3)
so the numbers are 19 +sqrt(3) & 19 - sqrt(3)..............................ans…

x + y = 38
xy = 358
=>
x (38 - x) = 358
x^2 -38x + 358 = 0
=> x =19+√3, x = 19 - √3
y = 19 - √3, y = 19 + √3

so the answer is 19+√3, 19-√3

17 and 21