I'm litteraly confounded as to how to turn this equation
y=x^2+18x-4
and I have no idea how to annihilate this inequality
-x^2+x+6 is greater than or equal to 0
1. Complete the square:
$\displaystyle y = x^2+18x-4 ~\iff~ y = x^2+18x + {\color{blue} 81 - 81 } - 4 ~\iff~ \boxed{y = (x+9)^2-85}$
Therefore the vertex is at V(-9, -85)
2. $\displaystyle -x^2+x+6 \geq 0 ~\iff~ x^2-x \leq 6$
Complete the square now;
$\displaystyle x^2-x + {\color{blue} \frac14 }\leq 6+ {\color{blue} \frac14 } ~\iff~ \left(x-\frac12 \right)^2 \leq \left(\frac52 \right)^2$
Calculate the square root on both sides:
$\displaystyle \left|x-\frac12 \right| \leq \frac52 ~\implies~ -\frac52 \leq x-\frac12 \leq \frac52$
Finally you get:
$\displaystyle \boxed{-2\leq x \leq 3}$