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6z^2-3z-30=0
a = 6; b = -3; c = -30;
Δ = b2-4ac
Δ = -32-4·6·(-30)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*6}=\frac{-24}{12} =-2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*6}=\frac{30}{12} =2+1/2 $
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