I would think it would be B
The height is 5 and 1/4 then the base is 2 and 1/2
You can convert these fractions into decimals and them multiply them to get your area.
<h2>1)</h2>

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

<h2>2)</h2>
Well he started off correct to the point of completing the square.
