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3z^2+20z-7=0
a = 3; b = 20; c = -7;
Δ = b2-4ac
Δ = 202-4·3·(-7)
Δ = 484
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{484}=22$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-22}{2*3}=\frac{-42}{6} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+22}{2*3}=\frac{2}{6} =1/3 $
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