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

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Solution for -(x+2)(x-4)=0 equation:



-(x+2)(x-4)=0
We multiply parentheses ..
-(+x^2-4x+2x-8)=0
We get rid of parentheses
-x^2+4x-2x+8=0
We add all the numbers together, and all the variables
-1x^2+2x+8=0
a = -1; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-1}=\frac{-8}{-2} =+4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-1}=\frac{4}{-2} =-2 $

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