2*x-ln(x)+0,5=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
/ / 1/2\\ / / 1/2\\
re\W\-2*e // I*im\W\-2*e //
x1 = - -------------- - ----------------
2 2
$$x_{1} = - \frac{\operatorname{re}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2} - \frac{i \operatorname{im}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2}$$
x1 = -re(LambertW(-2*exp(1/2)))/2 - i*im(LambertW(-2*exp(1/2)))/2
Sum and product of roots
[src]
/ / 1/2\\ / / 1/2\\
re\W\-2*e // I*im\W\-2*e //
- -------------- - ----------------
2 2
$$- \frac{\operatorname{re}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2} - \frac{i \operatorname{im}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2}$$
/ / 1/2\\ / / 1/2\\
re\W\-2*e // I*im\W\-2*e //
- -------------- - ----------------
2 2
$$- \frac{\operatorname{re}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2} - \frac{i \operatorname{im}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2}$$
/ / 1/2\\ / / 1/2\\
re\W\-2*e // I*im\W\-2*e //
- -------------- - ----------------
2 2
$$- \frac{\operatorname{re}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2} - \frac{i \operatorname{im}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2}$$
/ / 1/2\\ / / 1/2\\
re\W\-2*e // I*im\W\-2*e //
- -------------- - ----------------
2 2
$$- \frac{\operatorname{re}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2} - \frac{i \operatorname{im}{\left(W\left(- 2 e^{\frac{1}{2}}\right)\right)}}{2}$$
-re(LambertW(-2*exp(1/2)))/2 - i*im(LambertW(-2*exp(1/2)))/2
x1 = -0.268160645457627 + 0.9262953167672*i
x1 = -0.268160645457627 + 0.9262953167672*i