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

Solution for (x+2)(3-x)+6=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+x+12=0

a = -1; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-1)·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-1}=\frac{6}{-2}
=-3
$