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6z^2+11z+5=0
a = 6; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·6·5
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*6}=\frac{-12}{12} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*6}=\frac{-10}{12} =-5/6 $
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