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3z^2+71z+20=0
a = 3; b = 71; c = +20;
Δ = b2-4ac
Δ = 712-4·3·20
Δ = 4801
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(71)-\sqrt{4801}}{2*3}=\frac{-71-\sqrt{4801}}{6} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(71)+\sqrt{4801}}{2*3}=\frac{-71+\sqrt{4801}}{6} $
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