Hi,

x= first even integer (x+2)= second even integer

x^2 + (x+2)^2 = 20   |Nice work on the Set Up!

x^2 + x^2 + 4x + 4 = 20

2x^2 + 4x - 16 = 0

 x^2 + 2x - 8 = 0

(x+4)(x-2) = 0

(x+4) = 0 ,  x = -4

(x-2) = 0 ,  x = 2

Integers -4,-2   and 2,4

CHECKING our Answer***

 sum of squares = 20 

Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
 good so far

continue and solve the equation

x^2 + (x+2)^2 = 20

x^2 + (x+2)*(x+2) = 20

 (2x^2+4x+4) = 20

x^2+2x+2=10

x^2+2x-8=0

(x+4)*(x-2)=0



x=-4 

x+2=-2

and

 x=2

x+2=4

check

(-4)^2+(-2)^2=20

16+4=20

ok

2^2+4^2=20

4+16=20

ok





 

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