If $x+y=3$ and $\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$, then the value of $\left(x^2+y^2\right)$ is :

29 28 26 34

29

$x+y=3$ $\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$ \(\frac{y + x}{xy}\) = -\(\frac{3}{10}\)  \(\frac{3}{xy}\) = -\(\frac{3}{10}\)  xy = -10 ( a + b )2 = a2 + b2 + 2ab $\left(x^2+y^2\right)$ =  ( 3 )2 - 2 × -10 $\left(x^2+y^2\right)$ = 29