What is the general solution of a multivariate quadratic equation

If you want the general solution, first we treat one of the equations as quadratic in $x_1$ or $x_2$ and 'solve it'. Let's pick $x_1$ and the first equation:

$$\color{green}{x_1}=\frac{-(c_a+e_a \color{blue}{x_2}) \pm \sqrt{(c_a+e_a \color{blue}{x_2})^2-4a_a(b_a \color{blue}{x_2^2}+d_ax_2+f_a)}}{2a_a}$$

Then you have to substitue $x_1$ into the second equation. The best way would be to use the first equation again for $x_1^2$:

$$\color{green}{x_1^2}=-\frac{1}{a_a} \left((c_a+e_a \color{blue}{x_2})\color{green}{x_1}+b_a \color{blue}{x_2^2}+d_a\color{blue}{x_2}+f_a \right)$$

Now from the second equation you get:

$$-\frac{a_b}{a_a} \left((c_a+e_a \color{blue}{x_2})\color{green}{x_1}+b_a \color{blue}{x_2^2}+d_a\color{blue}{x_2}+f_a \right)+b_b\color{blue}{x_2^2}+c_b\color{green}{x_1}+d_b\color{blue}{x_2}+e_b \color{green}{x_1} \color{blue}{x_2}+f_b=0$$

Now let's separate $x_1$ so we can deal with the square root:

$$(a_b(c_a+e_a \color{blue}{x_2})-a_a e_b \color{blue}{x_2}-a_ac_b)\color{green}{x_1}=a_a(b_b \color{blue}{x_2^2}+d_b \color{blue}{x_2}+f_b)-a_b(b_a \color{blue}{x_2^2}+d_a\color{blue}{x_2}+f_a)$$

$$(a_bc_a-a_ac_b+(a_be_a-a_a e_b)\color{blue}{x_2})\color{green}{x_1}=(a_ab_b-a_bb_a)\color{blue}{x_2^2}+(a_ad_b-a_bd_a)\color{blue}{x_2}+a_af_b-a_bf_a$$

Now we substitue $x_1$ and get the square root on one side:

$$\pm \sqrt{(c_a+e_a \color{blue}{x_2})^2-4a_a(b_a \color{blue}{x_2^2}+d_a\color{blue}{x_2}+f_a)}(a_bc_a-a_ac_b+(a_be_a-a_a e_b)\color{blue}{x_2})= \\ =(a_bc_a-a_ac_b+(a_be_a-a_a e_b)\color{blue}{x_2})(c_a+e_a \color{blue}{x_2})+ \\ +2a_a\left((a_ab_b-a_bb_a)\color{blue}{x_2^2}+(a_ad_b-a_bd_a)\color{blue}{x_2}+a_af_b-a_bf_a\right)$$

Now you square both sides and since the maximum degree of $x_2$ right now is $2$, after squaring you get a quartic (4th degree) equation.

It has a general solution in radicals, but I certainly won't be writing it here.

Feeling hopeful about $n>2$ yet?

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Roots

Quadratics

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