Design Steps

V_{o u t} = V_{d d} \times \frac{R_{1}}{R_{N T C} + R_{1}} \times \frac{(R_{2} | | R_{3}) + R_{4}}{(R_{2} | | R_{3})} - (\frac{R_{4}}{R_{3}} \times V_{d d})

Calculate the value of R₁ to produce a linear output voltage. Use the minimum and maximum values of the NTC to obtain a range of values for R₁.
$R_{N T C M a x} = R_{N T C @ 25 C} = 2.252 k Ω, R_{N T C M i n} = R_{N T C @ 50 C} = 819.7 Ω$
$R_{1} =$ R N T C @ 25 C × R N T C @ 50 C = 2 . 252 k Ω × 819 . 7 Ω = 1 . 359 k Ω ≈ 1 . 37 k Ω
Calculate the input voltage range.
$V_{i n M i n} = V_{d d} \times \frac{R_{1}}{R_{N T C M a x} + R_{1}} = 3.3 V \times \frac{1.37 k Ω}{2.252 k Ω + 1.37 k Ω} = 1.248 V$
$V_{i n M a x} = V_{d d} \times \frac{R_{1}}{R_{N T C M i n} + R_{1}} = 3.3 V \times \frac{1.37 k Ω}{819.7 Ω + 1.37 k Ω} = 2.065 V$
Calculate the gain required to produce the maximum output swing.
$G_{i d e a l} = \frac{V_{o u t M a x} - V_{o u t M i n}}{V_{i n M a x} - V_{i n M i n}} = \frac{3.25 V - 0.05 V}{2.065 V - 1.248 V} = 3.917 \frac{V}{V}$
Solve for the parallel combination of R₂ and R₃ using the ideal gain. Select R₄= 1.5 kΩ (Standard Value).

${(R_{2} | | R_{3})}_{i d e a l} = \frac{R_{4}}{G_{i d e a l} - 1} = \frac{1.5 k Ω}{3.917}$ V V - 1 = 514 . 226 Ω
Calculate R₂ and R₃ based off of the transfer function and gain.

$R_{3} = \frac{R_{4} \times V_{d d}}{V_{i n M a x} \times G_{i d e a l} - V_{o u t M a x}} = \frac{1.5 k Ω \times 3.3 V}{2.065 V \times 3.917}$ V V - 3 . 25 V = 1023 . 02 Ω

$R_{2} = \frac{{(R_{2} | | R_{3})}_{i d e a l} \times R_{3}}{R_{3} - {(R_{2} | | R_{3})}_{i d e a l}} = \frac{514.226 Ω \times 1023.02 Ω}{1023.02 Ω - 514.226 Ω} = 1033.941 Ω$
Calculate the actual gain with the standard values of R₂ (1.02 kΩ) and R₃ (1.02 kΩ).
$G_{a c t u a l} = \frac{(R_{2} | | R_{3}) + R_{4}}{(R_{2} | | R_{3})} = \frac{510 Ω + 1.5 k Ω}{510 Ω} = 3.941 \frac{V}{V}$