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-35z^2+29z-6=0
a = -35; b = 29; c = -6;
Δ = b2-4ac
Δ = 292-4·(-35)·(-6)
Δ = 1
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{1}=1$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-1}{2*-35}=\frac{-30}{-70} =3/7 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+1}{2*-35}=\frac{-28}{-70} =2/5 $
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