[tex]\displaystyle\sum_{i=3}^{10}(2i-9)=2\sum_{i=3}^{10}i-9\sum_{i=3}^{10}1[/tex]
Recall that
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]
[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]
Your sums start at [tex]i=3[/tex], so in order to apply these formulas directly, you need to compensate for the missing first two terms:
[tex]\displaystyle\sum_{i=3}^{10}1=\sum_{i=1}^{10}1-(1+1)=10-2=8[/tex]
[tex]\displaystyle\sum_{i=3}^{10}i=\sum_{i=1}^{10}i-(1+2)=\frac{10\cdot11}2-3=52[/tex]
[tex]\implies\displaystyle\sum_{i=3}^{10}(2i-9)=2\cdot52-9\cdot8=\boxed{32}[/tex]
