(Z+2)(z-4)+(z-4)(z-6)=0

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Solution for (Z+2)(z-4)+(z-4)(z-6)=0 equation: 



(+2)(Z-4)+(Z-4)(Z-6)=0

We add all the numbers together, and all the variables

2(Z-4)+(Z-4)(Z-6)=0

We multiply parentheses

2Z+(Z-4)(Z-6)-8=0

We multiply parentheses ..

(+Z^2-6Z-4Z+24)+2Z-8=0

We get rid of parentheses

Z^2-6Z-4Z+2Z+24-8=0

We add all the numbers together, and all the variables

Z^2-8Z+16=0

a = 1; b = -8; c = +16;
Δ = b2-4ac
Δ = -82-4·1·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$Z=\frac{-b}{2a}=\frac{8}{2}=4$

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