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3z^2+19z+30=0
a = 3; b = 19; c = +30;
Δ = b2-4ac
Δ = 192-4·3·30
Δ = 1
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{1}=1$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*3}=\frac{-20}{6} =-3+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*3}=\frac{-18}{6} =-3 $
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