(7+5i)(2-8i)=1-i

Solution for (7+5i)(2-8i)=1-i equation:

(7+5i)(2-8i)=1-i

We move all terms to the left:

(7+5i)(2-8i)-(1-i)=0

We add all the numbers together, and all the variables

(5i+7)(-8i+2)-(-1i+1)=0

We get rid of parentheses

(5i+7)(-8i+2)+1i-1=0

We multiply parentheses ..

(-40i^2+10i-56i+14)+1i-1=0

We add all the numbers together, and all the variables

(-40i^2+10i-56i+14)+i-1=0

We get rid of parentheses

-40i^2+10i-56i+i+14-1=0

We add all the numbers together, and all the variables

-40i^2-45i+13=0

a = -40; b = -45; c = +13;
Δ = b2-4ac
Δ = -452-4·(-40)·13
Δ = 4105
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{-(-45)-\sqrt{4105}}{2*-40}=\frac{45-\sqrt{4105}}{-80}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{4105}}{2*-40}=\frac{45+\sqrt{4105}}{-80}
$