Calculating the roots of a quadratic with complex coefficients

Question

$x^2-(5i+14)x+2(5i+12)=0$

I got :

$\frac{(5i+14)+(75+100i)^{1/2}}{2}$

and

$\frac{(5i+14)-(75+100i)^{1/2}}{2}$

Wolfram gives : 2 and 12+5i

How do I reduce my solutions?

Answer 1

Here is a general approach: $$x^2-(5i+14)x=-10i-24$$ $$(x-2.5i-7)^2=-10i-24+(2.5i+7)^2$$ Let $z=x-2.5i-7$ We get: $z^2=18.75+25i$ after working out right hand side. Now this equation ought to be standard to solve. Backsub $z$ for $x$ gives you the 2 solutions.

Answer 2

$$x^2-(5i+14)x+2(5i+12)=0 \Longleftrightarrow$$ $$x^2-(14+5i)x+(24+10i)=0 \Longleftrightarrow$$ $$(x-(12+5i))(x-2)=0 \Longleftrightarrow$$ $$x-(12+5i)=0 \Longleftrightarrow \vee\space x-2=0 \Longleftrightarrow$$ $$x=12+5i \vee\space x=2 $$