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6z^2-29z+9=0
a = 6; b = -29; c = +9;
Δ = b2-4ac
Δ = -292-4·6·9
Δ = 625
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{625}=25$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-25}{2*6}=\frac{4}{12} =1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+25}{2*6}=\frac{54}{12} =4+1/2 $
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