Find the unique pair of real numbers (x,y) satisfying
(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4.
(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4
(3x^2 + 6x + 5)(y^2 + 4y + 5) = 2
So
3x^2 + 6x + 5 = 2
y^2 + 4y + 5 = 1
Solve for x and y. :)
=^._.^=
(-1, -2 ) via the quadratic formula (assigning each half of the left side = 2)