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-2z^2-3z+11=0
a = -2; b = -3; c = +11;
Δ = b2-4ac
Δ = -32-4·(-2)·11
Δ = 97
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{97}}{2*-2}=\frac{3-\sqrt{97}}{-4} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{97}}{2*-2}=\frac{3+\sqrt{97}}{-4} $
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