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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-9i^2+18i-8i+16)-(-1i+1)(i+1)=0

We get rid of parentheses

-9i^2+18i-8i-(-1i+1)(i+1)+16=0

We multiply parentheses ..

-9i^2-(-1i^2-1i+i+1)+18i-8i+16=0

We add all the numbers together, and all the variables

-9i^2-(-1i^2-1i+i+1)+10i+16=0

We get rid of parentheses

-9i^2+1i^2+1i-i+10i-1+16=0

We add all the numbers together, and all the variables

-8i^2+10i+15=0

a = -8; b = 10; c = +15;
Δ = b2-4ac
Δ = 102-4·(-8)·15
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{145}}{2*-8}=\frac{-10-2\sqrt{145}}{-16}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{145}}{2*-8}=\frac{-10+2\sqrt{145}}{-16}
$