One group of 20 observations has mean 8 and variance 5; another group of 30 observations has mean 13 and variance 10. The two groups are pooled, and…

The pooled mean is $\dfrac{20\cdot8+30\cdot13}{50}=11$. The pooled variance is $\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$ with $d_1=8-11=-3$ and $d_2=13-11=2$: $\dfrac{20(5+9)+30(10+4)}{50}=\dfrac{280+420}{50}=14$. Under $x\mapsto2x+3$ the variance is multiplied by $2^2=4$ and the shift of $3$ has no effect, so the new variance is $4\cdot14=56$. Forgetting the squared mean-differences gives $8$, and forgetting to square the factor $2$ gives $28$. Always compute the pooled variance from the pooled mean, not from the average of the two variances.