Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?
I recognize this as a quadratic equation, so I can use the quadratic formula to find the values of $d$ and $e$. The quadratic formula says that if $ax^{2}+bx+c=0$, then the solutions are $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=2$, $b=3$, and $c=-5$, so the solutions are $x=\frac{-3\pm\sqrt{3^{2}-4(2)(-5)}}{2(2)}$. Simplifying the expression, I get $x=\frac{-3\pm\sqrt{49}}{4}$. The square root of $49$ is $\pm 7$, so $x=\frac{-3\pm 7}{4}$. This means that $x=\frac{-3+7}{4}=1$ or $x=\frac{-3-7}{4}=-2.5$. I can choose either value to be $d$ and the other to be $e$, since the order does not matter. I will let $d=1$ and $e=-2.5$. Now I need to find the value of $(d-1)(e-1)$. Plugging in the values, I get $(1-1)(-2.5-1)=0(-3.5)=0$. 