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4z^2+19z-12=0
a = 4; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·4·(-12)
Δ = 553
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{553}}{2*4}=\frac{-19-\sqrt{553}}{8} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{553}}{2*4}=\frac{-19+\sqrt{553}}{8} $
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