The problem asks to solve the quadratic equation $2x^2 - 7x + 3 = 0$.

1. Problem Description

The problem asks to solve the quadratic equation $2x^2 - 7x + 3 = 0$.

2. Solution Steps

We can solve the quadratic equation $2x^2 - 7x + 3 = 0$ using factoring.

We look for two numbers that multiply to $2 \cdot 3 = 6$ and add up to $-7$. These numbers are $-6$ and $-1$.

Then we can rewrite the middle term as $-7x = -6x - x$.

So, the equation becomes $2x^2 - 6x - x + 3 = 0$.

Now we factor by grouping:

$2x(x - 3) - 1(x - 3) = 0$

$(2x - 1)(x - 3) = 0$

Setting each factor equal to zero, we have:

$2x - 1 = 0$ or $x - 3 = 0$

Solving for $x$ in each case:

$2x = 1$ implies $x = \frac{1}{2}$

$x = 3$

We can also use the quadratic formula:

For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by:

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

In our case, $a = 2$, $b = -7$, and $c = 3$.

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$

$x = \frac{7 \pm \sqrt{49 - 24}}{4}$

$x = \frac{7 \pm \sqrt{25}}{4}$

$x = \frac{7 \pm 5}{4}$

So, $x = \frac{7 + 5}{4} = \frac{12}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$

3. Final Answer

The solutions are $x = \frac{1}{2}$ and $x = 3$.