(-3-j)+(-1-3j)(0.707-0.707j)=0

Solution for (-3-j)+(-1-3j)(0.707-0.707j)=0 equation:

(-3-j)+(-1-3j)(0.707-0.707j)=0

We add all the numbers together, and all the variables

(-1j-3)+(-3j-1)(-0.707j+0.707)=0

We get rid of parentheses

-1j+(-3j-1)(-0.707j+0.707)-3=0

We multiply parentheses ..

(+0j^2-2.121j+0j-0.707)-1j-3=0

We get rid of parentheses

0j^2-2.121j+0j-1j-0.707-3=0

We add all the numbers together, and all the variables

j^2-2.121j-3.707=0

a = 1; b = -2.121; c = -3.707;
Δ = b2-4ac
Δ = -2.1212-4·1·(-3.707)
Δ = 19.326641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2.121)-\sqrt{19.326641}}{2*1}=\frac{2.121-\sqrt{19.326641}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.121)+\sqrt{19.326641}}{2*1}=\frac{2.121+\sqrt{19.326641}}{2}
$