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3z(3z+20-21)=0
We add all the numbers together, and all the variables
3z(3z-1)=0
We multiply parentheses
9z^2-3z=0
a = 9; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·9·0
Δ = 9
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{9}=3$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*9}=\frac{0}{18} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*9}=\frac{6}{18} =1/3 $
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