(x-19)(x+3)=0

Solution for (x-19)(x+3)=0 equation:

(x-19)(x+3)=0

We multiply parentheses ..

(+x^2+3x-19x-57)=0

We get rid of parentheses

x^2+3x-19x-57=0

We add all the numbers together, and all the variables

x^2-16x-57=0

a = 1; b = -16; c = -57;
Δ = b2-4ac
Δ = -162-4·1·(-57)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-22}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+22}{2*1}=\frac{38}{2}
=19
$