Find the x-intercepts of the quadratic function c(x) = 2x^2 + x - 10.

1. Understand x-intercepts: The x-intercepts of a function are the points where the graph of the function crosses the x-axis. At these points, the y-coordinate (or the function's value, $c(x)$) is always zero.

1. Set the function to zero: To find the x-intercepts, we need to solve the equation $c(x) = 0$. So, we set the given function equal to zero: $$2x^2 + x - 10 = 0$$ This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a=2$, $b=1$, and $c=-10$.

3. Solve the quadratic equation: We can solve this quadratic equation using the quadratic formula, which is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Substitute the values of $a$, $b$, and $c$ into the formula:
$$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-10)}}{2(2)}$$

4. Calculate the discriminant: First, calculate the value inside the square root (the discriminant):
$$1^2 - 4(2)(-10) = 1 - (-80) = 1 + 80 = 81$$

5. Continue solving for x: Now substitute the discriminant back into the quadratic formula:
$$x = \frac{-1 \pm \sqrt{81}}{4}$$
$$x = \frac{-1 \pm 9}{4}$$

6. Find the two x-intercepts: We have two possible values for $x$ because of the $\pm$ sign:
For the '+' sign:
$$x_1 = \frac{-1 + 9}{4} = \frac{8}{4} = 2$$
For the '-' sign:
$$x_2 = \frac{-1 - 9}{4} = \frac{-10}{4} = -\frac{5}{2}$$

1. Write the x-intercepts as coordinates: The x-intercepts are the x-values where $c(x)=0$. The y-coordinate at these points is 0. Therefore, the x-intercepts are $(2, 0)$ and $(-\frac{5}{2}, 0)$.