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The solutions or roots so-called are really just the x-intercepts, and for a quadratic equation, the vertex is found right half-way between those x-intercepts.
we can get the intercepts by zeroing out "y" or f(x), so let's take a peek at one of those,
so, the x-intercepts or solutions are at 5 and -1, let's take a peek what's the halfway point.
-1----0----1---- 2 ----3-----4------5
well then, now we know the vertex is at x = 2, but, what's the y-coordinate of it anyway?
y = (x - 5)(x + 1)
y = (2 - 5)(2 + 1)
y = (-3)(3)
y = -9
we can get the intercepts by zeroing out "y" or f(x), so let's take a peek at one of those,
so, the x-intercepts or solutions are at 5 and -1, let's take a peek what's the halfway point.
-1----0----1---- 2 ----3-----4------5
well then, now we know the vertex is at x = 2, but, what's the y-coordinate of it anyway?
y = (x - 5)(x + 1)
y = (2 - 5)(2 + 1)
y = (-3)(3)
y = -9