For what $x$ and $y$ polynomial has maximum value? For what $x,y\in\mathbb R$ does the polynomial
  $$-5x^2-2xy-2y^2+14x+10y-1$$
  attain a maximum?


My attempt:

I called $\alpha$ maximum value.

$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$

$$5x^2+2xy+2y^2-14x-10y+1+\alpha\geqslant 0$$
$$(x+y)^2+(y-5)^2+3x^2+(x-7)^2-73+\alpha\geqslant0$$
$$\alpha\geqslant73$$
So the lowest maximum value turned out to be $73$, but after checking answers I was wrong-maximum is $16$, so my further plans to calculate from that $x$ and $y$ seemed purposless. I'd like to see solution using only high school knowledge.

Ans: $x=1$, $y=2$
Write $$f(x)=-5x^2-(2y-14)x-2y^2+10y-1$$ This quadratic function on $x$ with parameter $y$ achieves maximum at 

$$p=-{b\over 2a} = {2y-14\over -10}$$ and this maximum is $$ q= -{b^2-4ac\over 4a} = {(2y-14)^2+20(-2y^2+10y-1)\over 20}=$$

So you need to find the maximum of $$g(y)= (2y-14)^2+20(-2y^2+10y-1)$$

$$ =-36y^2+144y +176$$

And now you can repeat the procedure which was done for $x$.

The maximum is at $y= -{144\over -72}=2$ (and thus $x=1$) and that maximum value is...
let $k$ be a maximal value.

Thus, for any $x$ and $y$ we have $$-5x^2-2xy-2y^2+14x+10y-1\leq k$$ or
$$5x^2+2(y-7)x+2y^2-10y+1+k\geq0,$$ which gives
$$(y-7)^2-5(2y^2-10y+1+k)\leq0$$ or
$$9y^2-36y-44+5k\geq0,$$ which gives $$18^2-9(-44+5k)\leq0,$$ which gives $$k\geq16.$$
The equality occurs for $$y=\frac{36}{2\cdot9}=2$$ and $$x=-\frac{2(2-7)}{2\cdot5}=1,$$ which says that a minimal value of $k$, 

for which the first inequality is true for any values of $x$ and $y$, it's $16$.