I want to solve the following system for $t_1 + t_2$.
$$ v_f=v_i + a(t_1-t_2) $$ $$x_f=x_i+v_i(t_1+t_2)+\frac{1}{2}a(t_1^2−t_2^2)+at_1t_2$$
I've tried solving for $t_1$ and substituting, but the equations get so complex that I don't trust myself to do every step correctly.
Here is the related question on Physics Stackexchange: How long does it take to optimally change position and velocity?
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See velocity-time diagram below, based on the original problem statement.
For the first part of the area under the curve, $s_0=\frac12(v_i+v_f)(t_1-t_2) $. Also, $v_f=v_i+a(t_1-t_2)$. Hence $$s_0=\frac 12 (v_f+v_i)\frac{v_f-v_i}a=\frac{v_f^2-v_i^2}{2a}$$
The remaining part of the area under the curve is given by $$v_f(2t_2)+\frac 12 (2t_2)(v_f+at_2)=at_2^2+3v_ft_2$$
The total area under the curve is the total distance travelled, i.e. $x_f-x_i$, hence
$$\begin{align} s_0+at_2^2+3v_ft_2&=x_f-x_i\\ \frac{v_f^2-v_i^2}{2a}+at_2^2+3v_ft_2&=x_f-x_i\\ at_2^2+3v_ft_2+\frac{v_f^2-v_i^2}{2a}-(x_f-x_i)&=0\\ t_2&=\frac{-3v_f+\sqrt{9v_f^2-4a\left(\frac{v_f^2-v_i^2}{2a}-(x_f-x_i)\right)}}{2a} \qquad \text{as $t_2>0$}\\ &=\frac{-3v_f+\sqrt{7v_f^2+2v_i^2+4a(x_f-x_i)}}{2a}\\ t_1&=\frac{v_f-v_i}a+t_2\\ &=-\frac{v_i}a+\frac{-v_f+\sqrt{7v_f^2+2v_i^2+4a(x_f-x_i)}}{2a}\\ t_1+t_2&=\frac{-v_i-2v_f+\sqrt{7v_f^2+2v_i^2+4a(x_f-x_i)}}a \end{align}$$
It does look quite messy. How does it compare with your answer?
Hint: Observe that $t_1t_2 = \dfrac{(t_1+t_2)^2-(t_1-t_2)^2}{2}$. Thus this allows you to substitute $u = t_1+t_2, v= t_1-t_2$. Can you continue?