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3z^2+17z+20=0
a = 3; b = 17; c = +20;
Δ = b2-4ac
Δ = 172-4·3·20
Δ = 49
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{49}=7$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*3}=\frac{-24}{6} =-4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*3}=\frac{-10}{6} =-1+2/3 $
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