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4+5z-6z^2=0
a = -6; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-6)·4
Δ = 121
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{121}=11$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-6}=\frac{-16}{-12} =1+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-6}=\frac{6}{-12} =-1/2 $
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