(x+4)(3-x)=6

Solution for (x+4)(3-x)=6 equation:

(x+4)(3-x)=6

We move all terms to the left:

(x+4)(3-x)-(6)=0

We add all the numbers together, and all the variables

(x+4)(-1x+3)-6=0

We multiply parentheses ..

(-1x^2+3x-4x+12)-6=0

We get rid of parentheses

-1x^2+3x-4x+12-6=0

We add all the numbers together, and all the variables

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

a = -1; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-1}=\frac{6}{-2}
=-3
$