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5z(4z+3)=96
We move all terms to the left:
5z(4z+3)-(96)=0
We multiply parentheses
20z^2+15z-96=0
a = 20; b = 15; c = -96;
Δ = b2-4ac
Δ = 152-4·20·(-96)
Δ = 7905
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{-(15)-\sqrt{7905}}{2*20}=\frac{-15-\sqrt{7905}}{40} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{7905}}{2*20}=\frac{-15+\sqrt{7905}}{40} $
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