19=x^2+2x+4

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

19=x^2+2x+4

We move all terms to the left:

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

We get rid of parentheses

-x^2-2x-4+19=0

We add all the numbers together, and all the variables

-1x^2-2x+15=0

a = -1; b = -2; c = +15;
Δ = b2-4ac
Δ = -22-4·(-1)·15
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*-1}=\frac{10}{-2}
=-5
$