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(4/5)z=16
We move all terms to the left:
(4/5)z-(16)=0
Domain of the equation: 5)z!=0We add all the numbers together, and all the variables
z!=0/1
z!=0
z∈R
(+4/5)z-16=0
We multiply parentheses
4z^2-16=0
a = 4; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·4·(-16)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*4}=\frac{-16}{8} =-2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*4}=\frac{16}{8} =2 $
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