Answer :

first set up equation
x times y=xy

so
xy=-29
and
x+y=1
subtract x from both sides
y=1-x
subsitute 1-x for y in first equation
x(1-x)=-29
distribute
x-x^2=-29
add x^2 to both sides
x=-29+x^2
subtract x from both sides
0=x^2-x-29
so we can use the quadratic formula to solve for x if the equation=0 and it is in ax^2+bx+c form so

if
ax+bx+c=0 then x=[tex] \frac{ -b+/-\sqrt{b^2-4ac} }{2a} [/tex] that means x=[tex]\frac{ -b-\sqrt{b^2-4ac} }{2a}[/tex] or x=[tex] \frac{ -b-\sqrt{b^2-4ac} }{2a} [/tex] so


x^2-x-29
a=1
b=-1
c=-29
[tex]\frac{ -(-1)-\sqrt{-1^2-4(1)(-29)} }{2(1)}[/tex]=[tex]\frac{ +1-\sqrt{1^2-(-116)} }{2(1)}=\frac{ +1-\sqrt{1^2+116} }{2}=\frac{ +1-\sqrt{117} }{2}= \frac{1-10.816653826392}{2} = [tex] \frac{-9.816653826392}{2}= -4.908326913196[/tex]



the second number is
[tex]\frac{ -(-1)+\sqrt{-1^2-4(1)(-29)} }{2(1)}=\frac{ +1+\sqrt{1^2+(-116)} }{2(1)}=
\frac{1+10.816653826392}{2} = \frac{ +1+\sqrt{117} }{2}=
\frac{10.816653826392}{2}=5.908326913196[/tex]


the two numbers are
5.908326913196 and
-4.908326913196


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