z^2+63-16z=0

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Solution for z^2+63-16z=0 equation:



z^2+63-16z=0
a = 1; b = -16; c = +63;
Δ = b2-4ac
Δ = -162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*1}=\frac{14}{2} =7 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*1}=\frac{18}{2} =9 $

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