(x^2+4x)+1=0

Solution for (x^2+4x)+1=0 equation:

(x^2+4x)+1=0

We get rid of parentheses

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

The end solution:

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