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3z^2+17z+16=0
a = 3; b = 17; c = +16;
Δ = b2-4ac
Δ = 172-4·3·16
Δ = 97
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{-(17)-\sqrt{97}}{2*3}=\frac{-17-\sqrt{97}}{6} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{97}}{2*3}=\frac{-17+\sqrt{97}}{6} $
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