For what values of h does the equation have real solutions. √(3x-h)= x+3

Square both side to get

3x-h = (x+3)^2

distribute the squared

3x - h = x^2 + 6x + 9

subtract 3x from both sides

-h = x^2 + 3x + 9

The answer will only have real solutions if the right side is factorable

h= -19, and 2.25 are both values that would make the right side factorable

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.