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4z^2-24+10z=0
a = 4; b = 10; c = -24;
Δ = b2-4ac
Δ = 102-4·4·(-24)
Δ = 484
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{484}=22$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*4}=\frac{-32}{8} =-4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*4}=\frac{12}{8} =1+1/2 $
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