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

Solution for (x+2)(x-19)=0 equation:

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

We multiply parentheses ..

(+x^2-19x+2x-38)=0

We get rid of parentheses

x^2-19x+2x-38=0

We add all the numbers together, and all the variables

x^2-17x-38=0

a = 1; b = -17; c = -38;
Δ = b2-4ac
Δ = -172-4·1·(-38)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-21}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+21}{2*1}=\frac{38}{2}
=19
$