Nonlinear functions SAT question

The correct answer is $\frac{25}{4}$. The given equation can be rewritten in the form $f(x)=a{(x-h)}^2+k$, where $a$, $h$, and $k$ are constants. When $a>0$, $h$ is the value of $x$ for which $f(x)$ reaches its minimum. The given equation can be rewritten as $f(x)=4(x^2-\frac{50}{4}x)+126$, which is equivalent to $f(x)=4(x^2-\frac{50}{4}x+{(\frac{50}{8})}^2-{(\frac{50}{8})}^2)+126$. This equation can be rewritten as $f(x)=4({(x-\frac{50}{8})}^2-{(\frac{50}{8})}^2)+126$, or $f(x)=4{(x-\frac{50}{8})}^2-4{(\frac{50}{8})}^2+126$, which is equivalent to $f(x)=4{(x-\frac{25}{4})}^2-\frac{121}{4}$. Therefore, $h=\frac{25}{4}$, so the value of $x$ for which $f(x)$ reaches its minimum is $\frac{25}{4}$. Note that 25/4 and 6.25 are examples of ways to enter a correct answer.