X(x+1) = 30
=> x^2 + x - 30 = 0
=> ( x - 5 ) ( x + 6 ) = 0
1) Since the degree of equation is 2 , therefore it represents a parabola .
2) Since co-efficient of x^2 is positive so the parabola is concave (open) upwards .
3) Since its discriminant ( D = b^2 - 4ac ) is positive and non-zero, therfore the parabola intersects x-axis at two points = zeroes of the equation = 5 and -6
4) The vertex of the parbola is given at ( -b/2a , -D/4a )
=> Vertex is at ( -1/2 , -30.25 )