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3z^2-13z+12=0
a = 3; b = -13; c = +12;
Δ = b2-4ac
Δ = -132-4·3·12
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*3}=\frac{8}{6} =1+1/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*3}=\frac{18}{6} =3 $
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