Find the value of y if the points (-7,y) and (4,-3) are √130 units apart.

Thank you:)

(-7,Y), (4,-3). d = sqrt(130).

d^2 = (4-(-7))^2 + (-3-Y)^2 = 130,
121 + 9 + 6y + y^2 = 130,
130 + 6Y + Y^2 - 130 = 0,
Y^2 + 6Y = 0,
Y(Y + 6) = 0,
Y = 0,
Y + 6 = 0,
Y = -6.

Solution set: Y = 0, and -6.