(4+i)(8+2i)=0

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Solution for (4+i)(8+2i)=0 equation:



(4+i)(8+2i)=0
We add all the numbers together, and all the variables
(i+4)(2i+8)=0
We multiply parentheses ..
(+2i^2+8i+8i+32)=0
We get rid of parentheses
2i^2+8i+8i+32=0
We add all the numbers together, and all the variables
2i^2+16i+32=0
a = 2; b = 16; c = +32;
Δ = b2-4ac
Δ = 162-4·2·32
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:
$i=\frac{-b}{2a}=\frac{-16}{4}=-4$

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