z^2-16z+48=0

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Solution for z^2-16z+48=0 equation: 



z^2-16z+48=0

a = 1; b = -16; c = +48;
Δ = b2-4ac
Δ = -162-4·1·48
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*1}=\frac{8}{2}
=4
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*1}=\frac{24}{2}
=12
$

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