Question 59561
Let x = the first number
Let x+2= the second number
(x)(x+2)=528 [Multiplying the consecutive even integers equals 528]
x^2+2x=528 [Solve for x using the quadratic formula]
x^2+2x-528=0 
.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
.
{{{x = (-(2) +- sqrt( (2)^2-(4)(1)(-528) ))/((2)(1)) }}} 
x=22 or x = -24 
.
Checking for x=22
(22)(22+2)=528
Checking for x=-24
(-24)(-24+2)=528