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3z^2+32z-11=0
a = 3; b = 32; c = -11;
Δ = b2-4ac
Δ = 322-4·3·(-11)
Δ = 1156
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{1156}=34$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-34}{2*3}=\frac{-66}{6} =-11 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+34}{2*3}=\frac{2}{6} =1/3 $
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