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3z^2-11z+10=0
a = 3; b = -11; c = +10;
Δ = b2-4ac
Δ = -112-4·3·10
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*3}=\frac{10}{6} =1+2/3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*3}=\frac{12}{6} =2 $
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