The formula looks like this: $$ \bigl( a x^2 + b x + c \bigr)^2 = \; d $$
What I try to solve is the following system: \begin{align} d_1 &= (a_1 x^2 + b_1 x + c_1)^2 \tag{1} \\ d_2 &= (a_2 x^2 + b_2 x + c_2)^2 \tag{2} \\ 1 &= d_1 + d_2 \tag{3} \end{align} with unknowns $x$, $d_1$, and $d_2$. I need $x$.
I know I can solve this through a $4$th degree equation but I believe this might have a much easier solution.
This comes from a real problem. I need to find the collision time (t) of 2 balls with the same R, 2R = 1 the 2nd degree formula is : $$ d = t * (v - t * a / 2)$$
where: d - distance. $d^2=∆x^2+∆y^2$, v - speed at t0 = 0, a - acceleration (friction)
I know the initial speed, friction and the direction of each object. There should be only one solution for t, or none
This isn't a particularly complicated quartic to solve. Really unclear why you are introducing more variables to create a system of equations
First, let $d_-=-\sqrt d, d_+=\sqrt d$. Then, we have two solutions that are solutions to the quadratic $ax^2+bx+c=d_-$ and two that are solutions to $ax^2+bx+c=d_+$.
That in mind, we can apply the quadratic formula to the equations $ax^2+bx+c=\pm\sqrt{d}$ and get roots $\frac{-b\pm\sqrt{b^2-4ac\pm4a\sqrt{d}}}{2a}$