(8+9i)(2i)-(1-i)(1+i)=0

Solution for (8+9i)(2i)-(1-i)(1+i)=0 equation:

(8+9i)(2i)-(1-i)(1+i)=0

We add all the numbers together, and all the variables

(9i+8)2i-(-1i+1)(i+1)=0

We multiply parentheses

18i^2+16i-(-1i+1)(i+1)=0

We multiply parentheses ..

18i^2-(-1i^2-1i+i+1)+16i=0

We get rid of parentheses

18i^2+1i^2+1i-i+16i-1=0

We add all the numbers together, and all the variables

19i^2+16i-1=0

a = 19; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·19·(-1)
Δ = 332
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{83}}{2*19}=\frac{-16-2\sqrt{83}}{38}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{83}}{2*19}=\frac{-16+2\sqrt{83}}{38}
$