<s>[INST] Can't find error at completing the square I am desperatly looking for the mistake I did when completing the square. 
I have a function $f(x)=-4.905x^2+5x+6$
Nothing special. So when I was trying to find the peak of the curve I ran into a problem and couldn't figure out why this happens, since I have repeated the task about 5 times. 
I used the form: $(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$
When I solve left hand side to get the value of -0.509683996 which seems to fit for the x value of the peak. When I put in this value for x  into the original equation I also receive the correct value of about y=7.53 for the peak. 
But when I try to read of the the peak by using the RHS my equation seems to crash: I received after serial trials always something that is not equal to the y values of the vertex :
https://www.symbolab.com/solver/step-by-step/%5Cfrac%7B5%5E%7B2%7D%7D%7B-9.81%5E%7B2%7D%7D-%5Cfrac%7B6%7D%7B-4.905%7D
what happened? Why is my right hand side not equal to 7.53?
I must do something extremely wrong when trying to calculate right hand side and receive the y value of the vertex.
 [/INST] It seems that you are confusing finding the solutions of $ax^2+bx+c=0$ where $a\not=0$ with finding the vertex of the parabola $y=ax^2+bx+c$.
In order to find the solutions, we have
$$\begin{align}ax^2+bx+c=0&\implies x^2+\frac bax=-\frac ca\\\\&\implies x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\\\\&\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\\\\&\implies x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\\\&\implies x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}\end{align}$$
In order to find the vertex of the parabola $y=ax^2+bx+c$, we have
$$\begin{align}y&=ax^2+bx+c\\\\&=a\left(x^2+\frac bax\right)+c\\\\&=a\left(x^2+\frac bax+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)+c\\\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)+c\\\\&=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\end{align}$$
So, the vertex is
$$\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$$
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