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3z(3z+11)=12
We move all terms to the left:
3z(3z+11)-(12)=0
We multiply parentheses
9z^2+33z-12=0
a = 9; b = 33; c = -12;
Δ = b2-4ac
Δ = 332-4·9·(-12)
Δ = 1521
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{1521}=39$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-39}{2*9}=\frac{-72}{18} =-4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+39}{2*9}=\frac{6}{18} =1/3 $
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