7y^2+18y+22=6y^2+34y-41

Solve for y over the real numbers:

7 y^2 + 18 y + 22 = 6 y^2 + 34 y - 41

Subtract 6 y^2 + 34 y - 41 from both sides:

y^2 - 16 y + 63 = 0

The left hand side factors into a product with two terms:

(y - 9) (y - 7) = 0

Split into two equations:

y - 9 = 0 or y - 7 = 0

Add 9 to both sides:

y = 9 or y - 7 = 0

Add 7 to both sides:

Answer: y = 9 or y = 7