x^2+4x+5=19

Solution for x^2+4x+5=19 equation:

x^2+4x+5=19

We move all terms to the left:

x^2+4x+5-(19)=0

We add all the numbers together, and all the variables

x^2+4x-14=0

a = 1; b = 4; c = -14;
Δ = b2-4ac
Δ = 42-4·1·(-14)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{2}}{2*1}=\frac{-4-6\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{2}}{2*1}=\frac{-4+6\sqrt{2}}{2}
$