smc

From Vieta's Formulas, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$ in the original quadratic. The sum of the roots in the new quadratic is $$ar+b+as+b$$ $$a(r+s)+2b$$ $$b$$ The product of the roots in the new quadratic is $$(ar+b)(as+b)$$ $$a^{2}rs+arb+asb+b^2$$ $$a^{2}rs+ab(r+s)+b^2$$ $$ac-b^2+b^2$$ $$ac$$ Thus, the new quadratic is $x^2-bx+ac$. The answer is $\boxed{x^2-bx+ac=0}$.