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2z^2+11z-6=0
a = 2; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·2·(-6)
Δ = 169
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{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*2}=\frac{-24}{4} =-6 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*2}=\frac{2}{4} =1/2 $
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