(x-2)(x+3)=x+19

Solution for (x-2)(x+3)=x+19 equation:

(x-2)(x+3)=x+19

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

(+x^2+3x-2x-6)-x-19=0

We add all the numbers together, and all the variables

(+x^2+3x-2x-6)-1x-19=0

We get rid of parentheses

x^2+3x-2x-1x-6-19=0

We add all the numbers together, and all the variables

x^2-25=0

a = 1; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·1·(-25)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*1}=\frac{10}{2}
=5
$