(4+(i^2))-(8+4i)+5=0

Solution for (4+(i^2))-(8+4i)+5=0 equation:

(4+(i^2))-(8+4i)+5=0

We add all the numbers together, and all the variables

(4+i^2)-(4i+8)+5=0

We get rid of parentheses

i^2-4i+4-8+5=0

We add all the numbers together, and all the variables

i^2-4i+1=0

a = 1; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·1·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{3}}{2*1}=\frac{4-2\sqrt{3}}{2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{3}}{2*1}=\frac{4+2\sqrt{3}}{2}
$