0=z^2+4z+1

Solution for 0=z^2+4z+1 equation:

0=z^2+4z+1

We move all terms to the left:

0-(z^2+4z+1)=0

We add all the numbers together, and all the variables

-(z^2+4z+1)=0

We get rid of parentheses

-z^2-4z-1=0

We add all the numbers together, and all the variables

-1z^2-4z-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:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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