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When we use the quadratic formula

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},$$

the entire numerator $-b\pm\sqrt{b^{2}-4ac}$ is divided by the denominator $2a$.
For the equation

$$2x^{2}-8x+6=0,$$

the coefficients are

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

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Plugging the numbers in

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

The denominator is $2a = 2\cdot 2 = 4$.
The numerator becomes

$$8\pm\sqrt{64-48}=8\pm\sqrt{16}=8\pm4 .$$

So we have

$$x=\frac{8\pm4}{4}.$$

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Why we “divide by 4”

The fraction $\dfrac{8\pm4}{4}$ means both terms in the numerator are each divided by the same denominator 4:

$$\frac{8+4}{4}= \frac{12}{4}=3,\qquad \frac{8-4}{4}= \frac{4}{4}=1.$$

You could also think of it as factoring a 4 out of the numerator first:

$$\frac{8\pm4}{4}= \frac{4(2\pm1)}{4}=2\pm1,$$

and then the 4’s cancel, leaving the same two results $3$ and $1$.

So the “division by 2 first” you noticed is actually the denominator $2a = 4$; it applies to the whole numerator, not just one part of it. This step ensures we correctly solve for the roots of the quadratic.

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Does that clear up the division step? Let me know if you’d like to see another example or a different way to simplify the fraction.