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

Solution for (4-x)(x+1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1x^2-1x+4x+4)=0

We get rid of parentheses

-1x^2-1x+4x+4=0

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-1)·4
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{-(3)-5}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-1}=\frac{2}{-2}
=-1
$