An algebra problem by anshu garg

The minimum value can be easily determined by completing the square:

$2x^2 - 3x + 4 = 2( x^2 - \frac {3}{2} x ) + 4 = 2(x - \frac {3}{4} )^2 - 2 × ( \frac {3}{4} )^2 + 4 =$

$= 2(x - \frac {3}{4} )^2 - \frac {9}{8} + 4 = 2(x - \frac {3}{4} )^2 + \frac {23}{8}$

The latter expression is minimal, when the square is zero ( x = 0.75 ).

Hence, the minimum value is:

$\boxed { \frac {23}{8} }$

The minimum value is given by $\frac{-D}{4a}$ using the graph

= $\frac{4ac-b^2}{4a}$

= $\frac{32-9}{8}$

= $\frac{23}{8}$