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3z^2+17z-6=0
a = 3; b = 17; c = -6;
Δ = b2-4ac
Δ = 172-4·3·(-6)
Δ = 361
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{361}=19$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*3}=\frac{-36}{6} =-6 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*3}=\frac{2}{6} =1/3 $
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