We need to find the roots of the equation $2(x+19) = 13x + 2(x^2-19)$.

1. Problem Description

We need to find the roots of the equation $2(x+19) = 13x + 2(x^2-19)$.

2. Solution Steps

First, we expand both sides of the equation:

$2(x+19) = 2x + 38$

$13x + 2(x^2-19) = 13x + 2x^2 - 38$

So, the equation becomes:

$2x + 38 = 13x + 2x^2 - 38$

Now, rearrange the equation to set it equal to zero:

$2x^2 + 13x - 2x - 38 - 38 = 0$

$2x^2 + 11x - 76 = 0$

We can solve this quadratic equation using the quadratic formula:

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

In our case, $a = 2$, $b = 11$, and $c = -76$. Plugging these values into the quadratic formula:

$x = \frac{-11 \pm \sqrt{11^2 - 4(2)(-76)}}{2(2)}$

$x = \frac{-11 \pm \sqrt{121 + 608}}{4}$

$x = \frac{-11 \pm \sqrt{729}}{4}$

$x = \frac{-11 \pm 27}{4}$

So, we have two possible solutions:

$x_1 = \frac{-11 + 27}{4} = \frac{16}{4} = 4$

$x_2 = \frac{-11 - 27}{4} = \frac{-38}{4} = -\frac{19}{2}$

3. Final Answer

$4, -\frac{19}{2}$