(3-4i)(1-i)-(2-i)(2+i)=0

Solution for (3-4i)(1-i)-(2-i)(2+i)=0 equation:

(3-4i)(1-i)-(2-i)(2+i)=0

We add all the numbers together, and all the variables

(-4i+3)(-1i+1)-(-1i+2)(i+2)=0

We multiply parentheses ..

(+4i^2-4i-3i+3)-(-1i+2)(i+2)=0

We get rid of parentheses

4i^2-4i-3i-(-1i+2)(i+2)+3=0

We multiply parentheses ..

4i^2-(-1i^2-2i+2i+4)-4i-3i+3=0

We add all the numbers together, and all the variables

4i^2-(-1i^2-2i+2i+4)-7i+3=0

We get rid of parentheses

4i^2+1i^2+2i-2i-7i-4+3=0

We add all the numbers together, and all the variables

5i^2-7i-1=0

a = 5; b = -7; c = -1;
Δ = b2-4ac
Δ = -72-4·5·(-1)
Δ = 69
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{69}}{2*5}=\frac{7-\sqrt{69}}{10}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{69}}{2*5}=\frac{7+\sqrt{69}}{10}
$