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3z^2-16z+21=0
a = 3; b = -16; c = +21;
Δ = b2-4ac
Δ = -162-4·3·21
Δ = 4
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{4}=2$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*3}=\frac{14}{6} =2+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*3}=\frac{18}{6} =3 $
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