Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$? 
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
Solution 1
Using factoring: 
$2x^{2}+3x-5=0$
$(2x+5)(x-1)=0$
$x = -\frac{5}{2}$ or $x=1$
So $d$ and $e$ are $-\frac{5}{2}$ and $1$. 
Therefore the answer is $\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}$

Solution 2
We can use the sum and product of a quadratic (a.k.a Vieta):
$(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{\mathrm{(B)}\ 0}$


Solution 3
By inspection, we quickly note that $x=1$ is a solution to the equation, therefore the answer is
$(d-1)(e-1)=(1-1)(e-1)=\boxed{\mathrm{(B)}\ 0}$

Solution 4
The form $(d-1)(e-1)$ resembles the factored form for the quadratic, namely $(x-d)(x-e)$ based on the information given. Note putting in 1 for $x$ in the that quadratic immediately yields the desired expression. Thus, $2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}$