Find $k$ such that $2x^2 + 5x + k = 0$ has equal roots Anyone have a method for solving this?

The equation $2x^2 + 5x + k = 0$ has equal roots. Find the value of $k$.

 A: Hint:
A quadratic equations has two identical roots iff its discriminant equals zero, so
$$\Delta:=5^2-4\cdot2\cdot k=0\iff\ldots$$
Solve now.
A: One method goes like this:
Two equal roots means, that you can write $2x^2+5x+k=a(x-b)^2$ where $b$ is the root.
Multiply this out and compare the coefficients.
A: Pretty sure you know the quadratic formula.

The roots of $ax^2+bx+c = 0$ are given by the quadratic formula: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

In the formula, $b^2-4ac$ is referred to as the discriminant, shown with $\Delta$. If this value is greater than $0$, there’ll be two real solutions. If it is equal to $0$, there’ll be one real solution (or two repeated solutions). If it is less than $0$, there’ll be no real solution.
Given $2x^2+5x+k = 0$ has one root, we can use the second case.
$$\Delta = 0$$
$$b^2-4ac = 0$$
$$5^2-4(2)(k) = 0$$
$$25-8k = 0$$
$$25 = 8k \implies \boxed{k = \frac{25}{8}}$$
A: Vieta: 
1)$x^2= k/2$, or
$x^2=k/2.$
2)$-2x=5/2$; $x= -5/4.$
Combining 2) and 1):
$k= 25/8.$
https://en.m.wikipedia.org/wiki/Vieta%27s_formulas
