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6z^2+24z-126=0
a = 6; b = 24; c = -126;
Δ = b2-4ac
Δ = 242-4·6·(-126)
Δ = 3600
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{3600}=60$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-60}{2*6}=\frac{-84}{12} =-7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+60}{2*6}=\frac{36}{12} =3 $
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