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3z^2+23z+30=0
a = 3; b = 23; c = +30;
Δ = b2-4ac
Δ = 232-4·3·30
Δ = 169
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{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-13}{2*3}=\frac{-36}{6} =-6 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+13}{2*3}=\frac{-10}{6} =-1+2/3 $
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