+1904 \(x+y=13\)
\(x\times y = -30\)
Take the first equation and solve for y.
\(x+y=13\)
\(x+y-x=13-x\)
\(0+y=13-x\)
\(y=13-x\)
Take that equation you solved for y and sbtitute y in the second equation.
\(x\times(13-x)=-30\)
\(13x-x^2=-30\)
\(13x-x^2+30=-30+30\)
\(13x-x^2+30=0\)
\(-x^2+13x+30=0\)
\((-1)\times(-x^2+13x+30)=0\times(-1)\)
\(x^2-13x-30=0\)
\((x-15)(x+2)=0\)
\(x-15=0\) \(x+2=0\)
\(x-15+15=0+15\) \(x+2-2=0-2\)
\(x-0=15\) \(x+0=-2\)
\(x=15\) \(x=-2\)
Subsitute the two numbers in the two original equations to see if the two answers work.
\(15+(-2)=13\) \(15\times(-2)=-30\)
\(15-2=13\) \(15\times(-2)=-30\)
\(13=13\) \(-30=-30\)
Both numbers checks out which means the two numbers that add up to 13 and multiply to -30 are 15 and -2.
+118608 Hi, I will try to give you some insight into how to work this out easily.
what two numbers add to 13 but multiply to -30?
First what are the pairs of factors of 30 (for now I will ignore the minus sign)
1 and 30
2 and 15
3 and 10
4 no,
5 and 6
there is no number between 5 and 6 so that is the last pair!
Now they have to multiply to give -30. This is a negative number so one will have to be negative and one will be positive.
They also have to add to be 13 which is positive so the biggest one will be positive.
So now our posibilities become
-1 and +30
-2 and +15
-3 and +10
-5 and +6
Now they have to add to be 13
-1 and +30 = 29
-2 and +15 = 13
-3 and +10 = 7
-5 and +6 =1
So the answer has to be -2 and 15 
If there had been no answer it means that the answer would have to be irrational or complex :)
