1. Express quadratic equation into perfect square form

Express $\displaystyle y = 2 - 5x - 3x^2$ in the form $\displaystyle y = a - b(x + c)^2.$Hence find the maximum value of y and the corresponding value of x.
This is what I did but I can't seem to get the answer $\displaystyle y = \frac{49}{12} - 3(x + \frac{5}{6})^2$

$\displaystyle y = 2 - 5x - 3x^2$
$\displaystyle y = -3 (x^2 + \frac{5}{3}x - \frac{2}{3})$
$\displaystyle y = -3 [(x + \frac{5}{6})^2 - \frac{2}{3} + \frac{23}{56})]$
$\displaystyle y = -\frac{1}{12} - 3(x + \frac{5}{6})^2$

Is there a careless mistake somewhere that I can't seem to spot? I've been redoing this for at least four times and I can't seem to see where I've gone wrong ):
Any help would be greatly appreciated and thank you very much in advance.

2. Originally Posted by caramelcake
This is what I did but I can't seem to get the answer $\displaystyle y = \frac{49}{12} - 3(x + \frac{5}{6})^2$

$\displaystyle y = 2 - 5x - 3x^2$
$\displaystyle y = -3 (x^2 + \frac{5}{3}x - \frac{2}{3})$
$\displaystyle y = -3 [(x + \frac{5}{6})^2 - \frac{2}{3} + \frac{23}{56})]$
This line is wrong.
$\displaystyle y = -\frac{1}{12} - 3(x + \frac{5}{6})^2$

Is there a careless mistake somewhere that I can't seem to spot? I've been redoing this for at least four times and I can't seem to see where I've gone wrong ):
Any help would be greatly appreciated and thank you very much in advance.

$\displaystyle y = 2-3\left(x^2+\frac53 x\right)$

$\displaystyle y = 2-3\left(x^2+\frac53 x + \frac{25}{36} - \frac{25}{36} \right)$
Here I added the missing square and subtracted it at once so the equation didn't change.
$\displaystyle y = 2-3\left(x+\frac56 \right)^2 + \frac{25}{12}$

$\displaystyle y = \frac{49}{12}-3\left(x+\frac56 \right)^2$

3. You could also expand the RHS, and compare the coefficients:

$\displaystyle 2-5x-3x^2 = a-b(x+c)^2 = (a-bc^2)-(2bc)x-bx^2$.

Thus $\displaystyle b = 3$, and $\displaystyle 2bc = 5$ hence $\displaystyle b = \frac{5}{6}$, and $\displaystyle a-bc^2 = 2$, so $\displaystyle a = \frac{49}{12}$.