Question 289806
 {{{ V(x) = x^2-28x + 13 }}}
You can take the derivative and set it equal to zero to find extrema.
{{{dV/dx=2x-28=0}}}

{{{2x-28-0}}}

{{{2x=28}}}

{{{x=14}}}

Since V''(x)=2, then the extrema is a minimum. 
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You can also check it graphically.
{{{ graph( 300, 300, -7, 28, -200, 100, x^2-28x + 13 ) }}}