Let \(d\) and \(e\) denote the solutions of \(3x^2+10x-25=0.\) Find \((d-e)^2.\)
\(\quad (d - e)^2\\ = d^2 + e^2 - 2de\\ =(d+e)^2 - 4de\\ =\left(\dfrac{-10}{3}\right)^2 - 4\left(\dfrac{-25}{3}\right)\\ =\dfrac{100}{9} + \dfrac{100}{3}\\ =\dfrac{400}{9}\)
Thank you so much
