Given the equation
$$\sqrt{3 x + 1} = 9 - x$$
$$\sqrt{3 x + 1} = 9 - x$$
We raise the equation sides to 2-th degree
$$3 x + 1 = \left(9 - x\right)^{2}$$
$$3 x + 1 = x^{2} - 18 x + 81$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 21 x - 80 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 21$$
$$c = -80$$
, then
D = b^2 - 4 * a * c =
(21)^2 - 4 * (-1) * (-80) = 121
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 5$$
$$x_{2} = 16$$
Because
$$\sqrt{3 x + 1} = 9 - x$$
and
$$\sqrt{3 x + 1} \geq 0$$
then
$$9 - x \geq 0$$
or
$$x \leq 9$$
$$-\infty < x$$
The final answer:
$$x_{1} = 5$$