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6z=z2
We move all terms to the left:
6z-(z2)=0
We add all the numbers together, and all the variables
-1z^2+6z=0
a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
Δ = 36
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{36}=6$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-1}=\frac{-12}{-2} =+6 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2} =0 $
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