\(2x^2+3x=4\implies2x^2+3x-4=0\implies x=\dfrac{-3\pm\sqrt{41}}4\)
Let \(\alpha\) be the root with the positive square root and \(\beta\) the root with the negative square root. Then
\(\dfrac1\alpha=\dfrac4{-3+\sqrt{41}}\quad\text{and}\quad\dfrac1\beta=\dfrac4{-3-\sqrt{41}}\)
The simplest quadratic with these roots can be written as
\(\left(x-\dfrac1\alpha\right)\left(x-\dfrac1\beta\right)=x^2-\left(\dfrac1\alpha+\dfrac1\beta\right)x+\dfrac1{\alpha\beta}=x^2-\dfrac34x-\dfrac12\)
