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sqrt(3-x)=x+1 equation

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Numerical solution:

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The solution

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  _______        
\/ 3 - x  = x + 1
$$\sqrt{3 - x} = x + 1$$
Detail solution
Given the equation
$$\sqrt{3 - x} = x + 1$$
$$\sqrt{3 - x} = x + 1$$
We raise the equation sides to 2-th degree
$$3 - x = \left(x + 1\right)^{2}$$
$$3 - x = x^{2} + 2 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} - 3 x + 2 = 0$$
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
$$a = -1$$
$$b = -3$$
$$c = 2$$
, then
D = b^2 - 4 * a * c = 

(-3)^2 - 4 * (-1) * (2) = 17

Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = - \frac{\sqrt{17}}{2} - \frac{3}{2}$$
$$x_{2} = - \frac{3}{2} + \frac{\sqrt{17}}{2}$$

Because
$$\sqrt{3 - x} = x + 1$$
and
$$\sqrt{3 - x} \geq 0$$
then
$$x + 1 \geq 0$$
or
$$-1 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = - \frac{3}{2} + \frac{\sqrt{17}}{2}$$
The graph
Rapid solution [src]
             ____
       3   \/ 17 
x1 = - - + ------
       2     2   
$$x_{1} = - \frac{3}{2} + \frac{\sqrt{17}}{2}$$
x1 = -3/2 + sqrt(17)/2
Sum and product of roots [src]
sum
        ____
  3   \/ 17 
- - + ------
  2     2   
$$- \frac{3}{2} + \frac{\sqrt{17}}{2}$$
=
        ____
  3   \/ 17 
- - + ------
  2     2   
$$- \frac{3}{2} + \frac{\sqrt{17}}{2}$$
product
        ____
  3   \/ 17 
- - + ------
  2     2   
$$- \frac{3}{2} + \frac{\sqrt{17}}{2}$$
=
        ____
  3   \/ 17 
- - + ------
  2     2   
$$- \frac{3}{2} + \frac{\sqrt{17}}{2}$$
-3/2 + sqrt(17)/2
Numerical answer [src]
x1 = 0.56155281280883
x1 = 0.56155281280883