We are asked to solve for $x$ in the equation $2x^2 - 7x + 5 = \frac{x^2+3x-4}{2}$.

1. Problem Description

We are asked to solve for $x$ in the equation $2x^2 - 7x + 5 = \frac{x^2+3x-4}{2}$.

2. Solution Steps

First, multiply both sides of the equation by 2 to eliminate the fraction:

$2(2x^2 - 7x + 5) = x^2 + 3x - 4$

$4x^2 - 14x + 10 = x^2 + 3x - 4$

Next, move all terms to the left side of the equation:

$4x^2 - x^2 - 14x - 3x + 10 + 4 = 0$

$3x^2 - 17x + 14 = 0$

Now, we can solve this quadratic equation using the quadratic formula or by factoring. Let's try factoring:

We are looking for two numbers that multiply to $3 \cdot 14 = 42$ and add up to $-17$. Those two numbers are $-3$ and $-14$.

So, we can rewrite the middle term as $-3x - 14x$:

$3x^2 - 3x - 14x + 14 = 0$

Now, factor by grouping:

$3x(x - 1) - 14(x - 1) = 0$

$(3x - 14)(x - 1) = 0$

Setting each factor equal to zero, we get:

$3x - 14 = 0$ or $x - 1 = 0$

$3x = 14$ or $x = 1$

$x = \frac{14}{3}$ or $x = 1$

3. Final Answer

The solutions for $x$ are $x = 1$ and $x = \frac{14}{3}$.