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9z^2-3z=20
We move all terms to the left:
9z^2-3z-(20)=0
a = 9; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·9·(-20)
Δ = 729
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{729}=27$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*9}=\frac{-24}{18} =-1+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*9}=\frac{30}{18} =1+2/3 $
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