Find the coordinates of intersection of a line and a circle 
There is a circle with a radius of $25$ ft and origin at $(0, 0)$ and a line segment from (0, -31) to (-37, 8).  Find the intersections of the line and circle.

I am asking for somebody to analyze what I am doing wrong in calculating the answer, given the question above in its exact format.  I almost never get an answer correct, and I would like an explanation as to why this appears to be the case.

I begin with finding the formula of the line

I used the slope formula, $\Delta y \over{\Delta x}$ to get ${-31 - 8 \over{0 - (-37)}} = {-39 \over{37}}$, which becomes $1.054$, so that $y = 1.054x + b$
$(0, -31)$ is the y-intercept when $x = 0$, so $y = 1.054x - 31$.  

Formula of circle

The format I learned is $(x - h)^2 + (y - k)^2 = r^2$.
Substitute known info: $(x - 0)^2 + (y - 0)^2 = 25$
Simplify: $x^2 + y^2 = 25$

Substitution

I took the circle's equation $x^2 + y^2 = 25$ and plugged in the value of y: $x^2 + (0.621x - 31)^2 = 25$.
I expanded and rounded it into $x^2 + 1.111x^2 + -65.348x + 961 = 25$.
It simplifies to $2.111x^2 - 65.348x + 936$.

Quadratic formula

The template: $x = {-b \pm \sqrt{b^2 - 4ac }\over{2a}}$ is used by plugging in values:
${65.348 \pm \sqrt{4270.361 - 4*2.111*936 }\over{2*2.111}} = {65.348 \pm \sqrt{4270.361 - 7903.584}\over{4.222}} = {65.348 \pm \sqrt{-3633.223}\over{4.222}}$.
Either I do not know what to do next or (as I suspect) this is the unfortunate answer to the problem.  The problem continues with questions about how long somebody is in the circle if they follow the line at a certain rate, which leads me to believe that there are real roots, but I certainly cannot find them.  Edit: slope and template errors corrected, answer is still incorrect.
 A: You made a mistake in citing the Quadratic Formula. The 'template' you need is:
$$x= \frac{-b\pm \sqrt{b^2 -4ac}}{2a}$$
Now you should be able to solve it.
A: Check your slope again: It is $-\frac{39}{37}\approx-1.05$. Check your circle, it is
$$x^2+y^2=625,$$
that is, $r^2=25^2=625$. I think you can get it from there, although I recommend you use fractions throughout until the end. For my final gift, here is a plot. If you hover over the intersection points, it will tell you what they are so you can verify your work.
A: The line equation is $y=(-39/37)x-31.$ To get the closest point on this line to the origin, intersect it with the perpendicular to it from $(0,0),$ which has equation $y=(37/39)x$. The nearest point on the line to $(0,0)$ is then seen to be $(a,b)$ where $a=-44733/2890,\ b=-42439/2890.$ The distance to the origin is then $\sqrt{a^2+b^2} \approx 21.336.$ Since this is less than the required radius of  $25$ there should be two points for the solution. [I could have sworn that initially the given radius was 5...]
