Solve for y where y is a real number y 2 Squarerooty 28 I

Solve for y, where y is a real number. y + 2 = Squareroot-y + 28 (If there is more than one solution, separate them with commas.) y =

y+2 = sqrt ( -y + 28)

squaring both sides to get rid of square root

(y+2)^2 = sqrt ( -y + 28) ^2

y^2 + 4y + 4 = -y+ 28

adding y and subtracting 28 from both sides

y^2 + 4y + 4 + y - 28 = 0

y^2 +5y - 24 = 0

(y+8)(y-3) = 0

y = -8

y = 3

plugging the value back into the equation to verify the solution

y+2 = sqrt ( -y + 28)

-8+2 = sqrt ( 8+28)

-6 = 6

hence y = -8 is not a solution

plugging y = 3

3+2 = sqrt ( -3+28)

5 = 5

only solution is y = 3