1. you're specifically asked to use it solve a quadratic equation ax²+bx+ac=0
2. you need to express a quadratic function y=ax²+bx+ac into a graph of the form: y = ±(x−p)²+q

Though the two cases look similar to the untrained eye, the perennial sin of treating the equation ax²+bx+ac=0 and the function y=ax²+bx+ac as the same thing has brought many students to grief (and blood-soaked test papers), especially after they anyhow bring terms left and right of the = sign when dealing with a function.

Obviously we're dealing with the second instant here as we need to find a, b and c in order to save Miss Loi's eyebrow, and we'll now attempt to dissect Ron's late night solution in a series of robust steps:

1. Step 1: Bring out coefficient of x² and put the x and x² terms inside brackets:

   y = −(x² − 3x) + 5

   The coefficient of x² is −1, so bringing it out will change the + 3x to − 3x inside the bracket (DUH! - yes simple algebra but then ...)

2. Step 2: + (half of x-coefficient)² − (x-coefficient)² within the brackets

   y = −(x² − 3x + (3/2)² − (3/2)²) + 5

3. Step 3: Move the new minus(−) term carefully out of the brackets, and be mindful of the need to multiply it with an constant outside the brackets.

   y = −(x² − 3x + (3/2)²) + (3/2)² + 5

   Note that the −(3/2)² has become a + (3/2)² because of the − sign outside the brackets.

   Tread carefully here, for this is careless mistake territory.

4. Step 4: Transform everything inside the brackets into a square term. Whatever constant that's outside the brackets, STAYS outside the brackets.

   y = −(x − (3/2))² + (3/2)² + 5

   Note that its (x − (3/2))² NOT (x + (3/2))² because of the negative − 3x within the brackets.

   You're deep, deep into careless mistakes territory now ...

5. Step 5: Sum up the constant terms outside the brackets to obtain the final expression y = ±(x−p)²+q, where (p, q) are the coordinates of the:

   • Min. point if (x−p)²+q (Recall the happy ∪ quadratic curve)
   • Max. point if −(x−p)²+q (Recall the sad ∩ quadratic curve )

   y = −(x − (3/2))² + 7.25

Miss Loi's eyebrow - fierce isn't it?

And so from our final expression above, the maximum (since there's a -ve sign outside the brackets) point of the graph is (3/2, 7.25)
⇒ c = 7.25
⇒ b = 3/2 = 1.5
To find the y-intercept a, simply equate x to 0 to get a=5.

N.B. Performing the 5 steps above is completely safe whether you're completing the square for a function or an equation, as there's no moving of terms left and right of the = sign.

Is this enough to tide you over miss loi?
cuts y-axis at y = 5, arch at x = 1.5, max value of y is at y=7.25.