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When we use the quadratic formula

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

the denominator is $2a$, not just the number 2.
For the equation

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

the coefficient $a$ is 2. Therefore

$$2a = 2\!\times\!2 = 4.$$

That is why the fraction ends up with a denominator of 4:

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

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Could we “divide by 2 first”?

Because every term in the quadratic $2x^{2}-8x+6$ shares a common factor of 2, we could simplify the equation before applying the formula:

$$\frac{2x^{2}-8x+6}{2}=0\quad\Longrightarrow\quad x^{2}-4x+3=0.$$

Now the coefficients are $a=1,\;b=-4,\;c=3$. Plugging these into the formula gives

$$x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(3)}}{2(1)} =\frac{4\pm\sqrt{16-12}}{2} =\frac{4\pm2}{2},$$

which also yields $x=3$ and $x=1$.

Both approaches are valid:

• Directly using the original coefficients gives a denominator of $2a=4$.
• Factoring out the common factor first reduces the numbers, giving a denominator of 2.

The result is the same; the choice is a matter of convenience.

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Final answer (either way):

$$x = 1 \quad \text{or} \quad x = 3$$