Complex number equation $x^2+x+(1-i)=0$

Factorize the equation as

$$x^2+x+(1-i)=x^2+x-i(1+i)=(x-i)(x+1+i)=0 $$

which yields the solutions $x=i,\>-(1+i)$.

For this particular quadratic equation, there is a very easy way to solve it: simple inspection suggests $x=i$ as a root of $x^2+x+(1-i)$, which is, indeed, easy to verify:

$$i^2+i+(1-i)=-1+i+(1-i)=0$$

The other root is easy to find from the fact that the sum of the roots of a quadratic is the negative of the coefficient of $x$, in this case $-1$, so the other root is $-1-i$. (As a doublecheck, note that the product of the roots, $i(-1-i)$, is equal to the constant term, $1-i$.)

To be a little fancy about what we've just done, the quadratic $x^2+x+(1-i)$ is a monic polynomial over the Gaussian integers, with prime constant coefficient $1-i$, so if it factors over the Gaussian integers, one of the roots must be a unit, i.e., one of $1$, $-1$, $i$, or $-i$.

Apparently you want to use the quadratic formula and are puzzled by the need to find the square root of $-3+4i$. The other answerers want to circumvent that question (which is fine to an extent). Let me show how to find that square root using the half-angle formulas.

We have $$ z=-3+4i=r(\cos\theta+i\sin\theta), $$ and easily calculate that the modulus $r=5$. We also observe that the number $z$ is in the second quadrant, so we can assume that $\theta\in(\pi/2,\pi)$. This implies that $\theta/2$ is in the first quadrant. Furhtermore, $5\cos\theta=-3$, so we can deduce that $\cos\theta=-3/5$.

The half-angle formulas then give that trig functions at $\theta/2$ are (both positive, because we saw that $\theta/2\in(0,\pi/2)$) $$ \begin{aligned} \cos\frac\theta2&=\sqrt{\frac{1+(-3/5)}2}=\sqrt{\frac15}=\frac1{\sqrt5},\\ \sin\frac\theta2&=\sqrt{\frac{1-(-3/5)}2}=\sqrt{\frac45}=\frac2{\sqrt5}. \end{aligned} $$ Therefore, for this choice of branch of the complex square root $$ \sqrt{z}=\sqrt{5}(\cos\frac\theta2+i\sin\frac\theta2)=1+2i. $$ It is, of course, easy to check this by squaring: $(1+2i)^2=-3+4i$.

Moral: You don't need the angle nor the half-angle. You only need their co/sines.