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3z^2+12z-420=0
a = 3; b = 12; c = -420;
Δ = b2-4ac
Δ = 122-4·3·(-420)
Δ = 5184
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{5184}=72$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-72}{2*3}=\frac{-84}{6} =-14 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+72}{2*3}=\frac{60}{6} =10 $
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