2) a) x² - 7x + 10 = 0; б) y² - 10y + 25 = 0; B) -t² + t + 3 = 0; г) 2a² - а = 3.

Решение:

1. a) \( x^2 - 7x + 10 = 0 \)
   
   \( D = (-7)^2 - 4 \cdot 1 \cdot 10 = 49 - 40 = 9 \)
   
   \( x_1 = \frac{7 + \sqrt{9}}{2} = \frac{7 + 3}{2} = 5 \)
   
   \( x_2 = \frac{7 - \sqrt{9}}{2} = \frac{7 - 3}{2} = 2 \)


2. б) \( y^2 - 10y + 25 = 0 \)
   
   \( D = (-10)^2 - 4 \cdot 1 \cdot 25 = 100 - 100 = 0 \)
   
   \( y = \frac{10}{2} = 5 \)


3. B) \( -t^2 + t + 3 = 0 \)
   
   \( t^2 - t - 3 = 0 \)
   
   \( D = (-1)^2 - 4 \cdot 1 \cdot (-3) = 1 + 12 = 13 \)
   
   \( t_1 = \frac{1 + \sqrt{13}}{2} \)
   
   \( t_2 = \frac{1 - \sqrt{13}}{2} \)


4. г) \( 2a^2 - a = 3 \)
   
   \( 2a^2 - a - 3 = 0 \)
   
   \( D = (-1)^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \)
   
   \( a_1 = \frac{1 + \sqrt{25}}{4} = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2} \)
   
   \( a_2 = \frac{1 - \sqrt{25}}{4} = \frac{1 - 5}{4} = \frac{-4}{4} = -1 \)

Ответ: а) \( 2; 5 \); б) \( 5 \); в) \( \frac{1 \pm \sqrt{13}}{2} \); г) \( \frac{3}{2}; -1 \).