[tex]2x^2-13x+20=0 \\ \\a=2, \ \ b=-13, \ c=20 \\ \\ \Delta =b^2-4ac =13^2-4\cdot 2\cdot 20 =169-160=9\\\\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{9}}{2\cdot 2 }=\frac{ 10}{4}=\frac{5}{2} =2.5\\\\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{9}}{2\cdot 2 }=\frac{ 13+3}{4}=\frac{16}{4}=4[/tex]
[tex]Answer : x=4 \ \ or \ \ x=2.5[/tex]
