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-4 x^2 -3x+2    is a dome shaped parabola        max will occur at x = -b/2a        b = -3   a = -4

   use this value of 'x' to calculate the max value

To put it into standard form (and to make it look better), multiply by -1. This is the resulting equation: \(-1(4x^2+3x-2)\). For the whole next part we will totally disregard the -1.Then we complete the square. But to do that we need to divide by four. \(4(x^2+\frac{3}{4}x-\frac{1}{2})\). Now we can easily complete the square. \(4[(x^2+\frac{3}{4}x+(\frac{3}{8})^2)-\frac{1}{2}]-\frac{9}{16}\). Then write it out into a simpler form: \(4[(x+\frac{3}{8})^2-\frac{1}{2}]-\frac{9}{16}\). Now expand what is within the parentheses. \(4(x+\frac{3}{8})^2-2-\frac{9}{16}\). Combine the constants \(\Rightarrow 4(x+\frac{3}{8})^2-\frac{41}{16}\).

But wait!!! If you remember what I said: we have to multiply by the negative one since we disregarded it back in the first paragraph. So we multiply the whole thing by negative one. This is what it will look like: \(-4(x+\frac{3}{8})^2+\frac{41}{16}\). By the Trivial Inequality, we know that that square right in the middle must be positive or 0. But if you look back at the equation: you have to multiply by a negative number (-4). So the smaller the square is the better. If the square can only be 0 or positive. We definitely want it to be 0. If you plug that into the equation, you will realize that it is only just \(\frac{41}{16}\). 

Edit: I realized I made 2 stupid mistakes! (Thanks EP!!!!)
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