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64=z2
We move all terms to the left:
64-(z2)=0
We add all the numbers together, and all the variables
-1z^2+64=0
a = -1; b = 0; c = +64;
Δ = b2-4ac
Δ = 02-4·(-1)·64
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-1}=\frac{-16}{-2} =+8 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-1}=\frac{16}{-2} =-8 $
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