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3z^2-3z-36=0
a = 3; b = -3; c = -36;
Δ = b2-4ac
Δ = -32-4·3·(-36)
Δ = 441
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{441}=21$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*3}=\frac{-18}{6} =-3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*3}=\frac{24}{6} =4 $
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