The equation $\sqrt{x^2+36 x+180}=2 x+15$ have two integer solutions are $x = -5$ and $x = -3$. How can I choose the integer numbers $a, b, c, d, t$ so that the equation $\sqrt{a x^2 + b x + c} = d x + t$ has two integer solutions? I tried

Solve[a x^2 + b x + c == (d x + t)^2, x]

{{x -> (-b + 2 d t - Sqrt[ b^2 - 4 a c + 4 c d^2 - 4 b d t + 4 a t^2])/( 2 (a - d^2))}, {x -> (-b + 2 d t + Sqrt[b^2 - 4 a c + 4 c d^2 - 4 b d t + 4 a t^2])/(2 (a - d^2))}}

How can I add conditions $(-b + 2 d t - Sqrt[ b^2 - 4 a c + 4 c d^2 - 4 b d t + 4 a t^2])/(2 (a - d^2))$ and $(-b + 2 d t + Sqrt[ b^2 - 4 a c + 4 c d^2 - 4 b d t + 4 a t^2])/(2 (a - d^2))$ are integer numbers.