Find the complex roots of the equation and do a check: $$ x^{2} + (2+i)x-1+7i=0 $$
What I've done: $$D=b^2-4ac=(2+i)^2-4(-1+7i)=4+4i+4-28i=8-24i$$ $$\sqrt{8-24i}=\pm(\sqrt{(640-8)/2}+i(640+8)/2) = \sqrt{316}+18i$$
Here the wrong calculation. And then it will be necessary to calculate $x_1$ and $x_2$.
All sorts of little errors.
$$D=(2+i)^2-4(-1+7i)=4+4i-1+4-28i=7-24i$$
You should now recognise $7, 24, 25$ as a pythagorean triple, or simply check that $$(7-24i)(7+24i)=25^2$$ whence (to use shorthand, as you have done) $\sqrt {7-24i}$ will be of the form $a+bi: a^2+b^2=25$ and you can spot $(4-3i)^2$
You've made a small mistake in the calculation of your Discriminant :
$$D=(2+i)^2 -4(-1+7i)=7-24i$$
Then, the roots will be :
$$x_{1,2} = \frac{-(2+i) \pm (7-24i)^{1/2}i }{2} = \frac{-2-i \pm [(-3i+4)^2]^{1/2}i}{2} $$
Can you take it from here ?
Multiplying with 4 we get $$(2x+2+i)^2 =7-24i=(4-3i)^2$$
so $$2x =-2-i\pm(4-3i)$$ so $x_1=1-2i$ and $x_2 = -3+i$.
This is a quadratic equation set to $0$, where $a=1$, $b=2+i$, and $c=-1+7i$. Plugging into the quadratic equation, we get:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$=\frac{-(2+i)\pm\sqrt{(2+i)^2-4(1)(-1+7i)}}{2(1)}$$
From this point it’s basic arithmetic which I will leave as an exercise to the reader.
