Answer
$$
\sqrt {x+1}=x^{2}-x
$$
the roots of the given equation, correct to six decimal places, are -0.484028 and 1.897179.
Work Step by Step
$$
\sqrt {x+1}=x^{2}-x
$$
We first rewrite the equation in standard form:
$$
\sqrt {x+1}-x^{2}+x=0
$$
Therefore we let
$$
f(x)=\sqrt {x+1}-x^{2}+x=0 .
$$
Then
$$
f^{'}(x)=\frac{1}{2 \sqrt{x+1}}-2 x+1
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{\sqrt{x_{n}+1}-x_{n}^{2}+x_{n}}{\frac{1}{2 \sqrt{x_{n}+1}}-2 x_{n}+1}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=\sqrt {x+1} $ and $y = x^{2}-x $ in the Figure .
(*) It appears that they intersect at a point whose x-coordinate is somewhat less than 2, so let’s take $x_{1} = 2$ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{\sqrt{x_{1}+1}-x_{1}^{2}+x_{1}}{\frac{1}{2 \sqrt{x_{1}+1}}-2 x_{1}+1}\\
&=(2)-\frac{\sqrt{(2)+1}-(2)^{2}+(2)}{\frac{1}{2 \sqrt{(2)+1}}-2 (2)+1}\approx 1.901174
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 1.897186,
x_{4}\approx 1.897179 \approx x_{5}
$$
(**) It appears that they intersect at a point whose x-coordinate is somewhat greater than -0.5, so let’s take $x_{1} = -0.5 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{\sqrt{x_{1}+1}-x_{1}^{2}+x_{1}}{\frac{1}{2 \sqrt{x_{1}+1}}-2 x_{1}+1}\\
&=(-0.5)-\frac{\sqrt{(-0.5)+1}-(-0.5)^{2}+(-0.5)}{\frac{1}{2 \sqrt{(-0.5)+1}}-2 (-0.5)+1} \approx -0.484155
\end{aligned}
$$
repeating we get
$$
x_{3}\approx -0.484028 \approx x_{4}
$$
we conclude that the roots of equation, correct to six decimal places, are -0.484028 and 1.897179.