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3z=1-4z^2
We move all terms to the left:
3z-(1-4z^2)=0
We get rid of parentheses
4z^2+3z-1=0
a = 4; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·4·(-1)
Δ = 25
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{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*4}=\frac{-8}{8} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*4}=\frac{2}{8} =1/4 $
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