(z-9)(z+7)=0

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Solution for (z-9)(z+7)=0 equation:



(z-9)(z+7)=0
We multiply parentheses ..
(+z^2+7z-9z-63)=0
We get rid of parentheses
z^2+7z-9z-63=0
We add all the numbers together, and all the variables
z^2-2z-63=0
a = 1; b = -2; c = -63;
Δ = b2-4ac
Δ = -22-4·1·(-63)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*1}=\frac{-14}{2} =-7 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*1}=\frac{18}{2} =9 $

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