Another cool thing I never mentioned that is cool about vertex form is that you can calculate x when the equation is in vertex form. For (x + 3)^2 – 1 = 0, you can say that x = 2 or 4, just by simply manipulating variables.
(x + 3)^2 – 1 = 0
(x + 3)^2 = 1 (add one to both sides)
x + 3 = 1 or 1 (take the ± square root of both sides; use 1 and 1 because 1
and 1 squared are both 1)
and 1 squared are both 1)
x = 2 or 4 (subtract three from both sides, from the 1 and the 1 to get
both answers)
both answers)
What about for standard form, using the coefficients a, b, and c. Maybe there is a formula we can derive just by following the steps involved in completing the square. Remember our original equation: ax^{2} + bx + c = 0. We’ll do it alongside an example utilizing every step, 2x^{2} – 4x + 6 = 0.
Starting Equation

ax^{2 }+ bx + c = 0

2x^{2} – 4x – 6 = 0

Move the constant over to the right hand side, just to make things less messy.

ax^{2} + bx = c

2x^{2} – 4x = 6

Divide through by a. Normally, we would factor it, but it is the same thing.

x^{2} + ^{b}/_{a}x = ^{c}/_{a}

x^{2} – 2x = 3

Take half of the new b term and square it.

(^{b}/_{2a})^{2} = ^{b^2}/_{4a^2}

(1)^{2} = 1

Add in and subtract out this number. For this, we’ll just add it to both sides, since c is on the other side.

x^{2} + ^{b}/_{a}x + ^{b^2}/_{4a^2} = ^{c}/_{a} + ^{b^2}/_{4a^2}

x^{2} – 2x + 1 = 3 + 1

Simplify the right hand side. For the first example, we will multiply the first fraction by ^{4a}/_{4a} to get a common denominator.

^{4a}/_{4a}(^{c}/_{a}) + ^{b^2}/_{4a^2}
^{4ac}/_{4a^2} + ^{b^2}/_{4a^2}
^{b^2 – 4ac}/_{4a^2}
x^{2} + ^{b}/_{a}x + ^{b^2}/_{4a^2} = ^{b^2 – 4ac}/_{4a^2}

x^{2} – 2x + 1 = 4

Convert the left hand side into squared form.

(x + ^{b}/_{2a})^{2} = ^{b^2 – 4ac}/_{4a^2}

(x – 1)^{2} = 4

Square root both sides. Don't forget to put our ± sign before the square root!

x + ^{b}/_{2a} = ±^{√(b^2 – 4ac)}/_{2a}

x – 1 = ±2

Isolate the x by adding/subtracting the remaining constant.

x = ^{b}/_{2a }± ^{√(b^2 – 4ac)}/_{2a}
x = ^{b ± √(b^2 – 4ac)}/_{2a}^{}

x = 1 ± 2
x = 1 or 3 
This formula is pretty wellknown, but I think it's cool how it is derived. This is a formula that you want to memorize, because it is so important.
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