This Is Why I Hate Beetroot

Let the roots be $\alpha, \beta, \gamma$ , and $\delta$ .

$\alpha + \beta = \gamma + \delta = t$ (say)

Now, use vieta's formulae:

$\alpha+\beta+\gamma+\delta = -a \neq 2$

$\Rightarrow t = \dfrac{-a}{2} \neq 1$

Now, as $a \in R$ , hence $t \in R$

Now, $\alpha \beta + \gamma \delta + (\alpha + \beta)(\gamma + \delta) = b$

$\Rightarrow t^2 + \alpha \beta + \gamma \delta = 1 - c$

Also,

$(\alpha \beta + \gamma\delta)(\alpha + \beta) = -c$

$\Rightarrow \alpha \beta + \gamma \delta = \dfrac{-c}{t}$

Putting this value in our previous equation,

$t^2 - \dfrac{c}{t} = 1 - c$

$\Rightarrow\dfrac{(t-1)(t^2+t+c)}{t} = 0$

Now, as $t \neq 1$ , $t^2+t+c$ must have real roots as $t \in R$ ,

So, $1- 4c \leq 0$