(z+2)(z+3)=0

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Solution for (z+2)(z+3)=0 equation:



(z+2)(z+3)=0
We multiply parentheses ..
(+z^2+3z+2z+6)=0
We get rid of parentheses
z^2+3z+2z+6=0
We add all the numbers together, and all the variables
z^2+5z+6=0
a = 1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*1}=\frac{-6}{2} =-3 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*1}=\frac{-4}{2} =-2 $

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