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16z^2-8z=3
We move all terms to the left:
16z^2-8z-(3)=0
a = 16; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·16·(-3)
Δ = 256
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{256}=16$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*16}=\frac{-8}{32} =-1/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*16}=\frac{24}{32} =3/4 $
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