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Step‑by‑step solution

1. Write the equation

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

2. Simplify (divide every term by the common factor 2)

   $$x^{2}-4x+3=0$$

3. Factor the quadratic – look for two numbers that multiply to $+3$ and add to $-4$.
   Those numbers are $-1$ and $-3$.

   $$x^{2}-4x+3=(x-1)(x-3)=0$$

4. Apply the zero‑product property (if a product is zero, at least one factor is zero).

   $$\begin{cases} x-1=0 \\ x-3=0 \end{cases}$$

5. Solve each simple equation

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

Check with the quadratic formula (optional)

For $ax^{2}+bx+c=0$, $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Here $a=2$, $b=-8$, $c=6$:

Thus $x=\dfrac{12}{4}=3$ or $x=\dfrac{4}{4}=1$, matching the factor‑method result.

Final Answer

If you’d like a visual walkthrough of these methods, the video below explains both factoring and the quadratic formula in detail.