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

Solution for (3-x)(x+1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-1}=\frac{2}{-2}
=-1
$