Question 1123447
<br>
In the quadratic equation<br>
ax^2+bx+c = 0<br>
the sum of the roots is -b/a and the product is c/a.<br>
This holds for quadratic equations with complex coefficients.<br>
With a=1 in this equation, the sum of the roots is -(2i-3) = 3-2i and the product is 5-i.<br>
Let the roots be p+qi and r+si.  Then their product is<br>
(pr-qs)+(ps+qr)i<br>
and their sum is<br>
(p+r)+(q+s)i<br>
So we have four equations in p, q, r, and s:<br>
pr-qs = 5; ps+qr = -1
and
p+r= 3; q+s = -2<br>
You can solve the problem by playing around with small integers, trying to find solutions for the second pair of equations that give correct results in the first pair of equations.<br>
I leave that to you as an exercise.<br>
Another way to solve the problem is by using the quadratic formula -- which also holds for quadratic equations with complex coefficients.<br>
Indeed, many students who are not comfortable with factoring techniques always go straight to the quadratic formula for solving any quadratic equation.<br>
For this equation, the quadratic formula gives us the two roots as<br>
{{{((3-2i) + sqrt((2i-3)^2-4(1)(5-1)))/2}}} and {{{((3-2i) - sqrt((2i-3)^2-4(1)(5-1)))/2}}}
{{{((3-2i) + sqrt((5-12i)-(20-4i))/2)}}} and {{{((3-2i) - sqrt((5-12i)-(20-4i)))/2}}}
{{{((3-2i) + sqrt(-15-8i))/2}}} and {{{((3-2i) - sqrt(-15-8i))/2}}}
{{{((3-2i) + (1-4i))/2}}} and {{{((3-2i) - (1-4i))/2}}}
{{{(2-3i)}}} and {{{(1+i)}}}<br>
Of course those are the two roots you should end up with if you try the trial-and-error method suggested above.<br>
So...<br>
Answer:  The two roots are (2-3i) and (1+i)