[tex]\bf \textit{that simply means }A=y\implies 50=x^2-9x+14
\\\\\\
0=x^2-9x+14-50\implies
\begin{array}{lccll}
x^2&-9x&-36\\
&\uparrow &\uparrow \\
&-12+3&-12\cdot 3
\end{array}
\\\\\\
0=(x-12)(x+3)\implies
\begin{cases}
0=x-12\implies &\boxed{12=x}\\
0=x+3\implies &-3=x
\end{cases}[/tex]
since the value for "x" cannot be negative, because is a length unit, so is not -3.
