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What the “2” in the denominator means

The quadratic formula

comes from completing the square on the general quadratic

When you finish the algebra, the term that multiplies the whole fraction is $2a$, not just the number 2.

Why we don’t “divide by 2 first”

1. Identify the coefficients
   For your equation $2x^{2}-8x+6=0$ we have

   $$a=2,\qquad b=-8,\qquad c=6 .$$

2. Plug them into the formula

   $$x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(6)}}{2(2)} .$$

   Notice the denominator is $2a = 2\times 2 = 4$.
   The “2” is multiplied by the coefficient $a$; it isn’t a separate step that you do before evaluating the numerator.

3. Simplify the numerator first

   $$-(-8)=8,\qquad \sqrt{(-8)^{2}-4(2)(6)}=\sqrt{64-48}= \sqrt{16}=4 .$$

   So the numerator becomes $8\pm4$, giving two possibilities:

   $$8+4=12\quad\text{or}\quad 8-4=4 .$$

4. Now divide each result by the whole denominator $2a=4$

   $$x_{1}= \frac{12}{4}=3,\qquad x_{2}= \frac{4}{4}=1 .$$

If you tried to “divide by 2 first,” you would be ignoring the factor $a$ (which is 2 in this problem) and you would not get the correct roots.

Quick recap

• The denominator $2a$ comes directly from the derivation of the quadratic formula.
• You first evaluate the entire numerator $(-b \pm \sqrt{b^{2}-4ac})$.
• Only after you have the two possible numerators do you divide each by the full denominator $2a$.

Does that clear up why the division happens the way it does? Let me know if any part is still fuzzy!