Answer: B. $-50$

Explanation:$1$. The given equation is a quadratic equation of the form $ax^2+bx+c\_{eq}=0$. In this equation, $a\_{coeff}=-2$, $b\_{coeff}=20$, and the constant term is $c$. (Using $a\_{coeff}$, $b\_{coeff}$ to avoid confusion with the 'c' in the equation itself).

$2$. For a quadratic equation to have exactly one solution (a repeated real root), its discriminant ($D$) must be equal to zero. The discriminant is given by $D = (b\_{coeff})^2 - 4(a\_{coeff})(c\_{eq})$. Here, $c\_{eq}$ is the constant 'c' from the problem statement.

$3$. Set up the discriminant equation and set it to zero: $D = (20)^2 - 4(-2)(c) = 0$

$4$. Solve for c: $(20)^2 = 400$ $-4(-2)(c) = 8c$ So, the equation becomes: $400 + 8c = 0$ Subtract $400$ from both sides: $8c = -400$ Divide by $8$: $c = \dfrac{-400}{8}$ $c = -50$

$5$. Check the options: Option [B] is $-50$, which matches our result.

$6$. Conclusion: The value of c for which the equation has exactly one solution is $-50$.