Roots

$kx^2+2x^2-3kx+k=0$

$(k+2)x^2-3kx+k=0$ $\color{#3D99F6}\large \implies$ It is a quadratic equation in the form of $ax^2+bx+c=0$

So, $a=k+2, b=-3k$ and $c=k$

So that the roots are equal, the discriminant must be equal to $0$ . That is, $b^2-4ac=0$

So, we have

$(-3k)^2-4(k+2)(k)=0$ $\color{#3D99F6}\large \implies$ $9k^2-4k(k+2)=0$ $\color{#3D99F6}\large \implies$ $9k^2-4k^2-8k=0$ $\color{#3D99F6}\large \implies$ $5k^2-8k=0$ $\color{#3D99F6}\large \implies$ $k(5k-8)=0$

$k=0$

$5k-8=0$ $\color{#3D99F6}\large \implies$ $5k=8$ $\color{#3D99F6}\large \implies$ $k=\dfrac{8}{5}$

Finally,

$0+\dfrac{8}{5}=$ $\color{#D61F06}\boxed{\large 1.6}$