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14z^2-3z-5=0
a = 14; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·14·(-5)
Δ = 289
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{289}=17$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-17}{2*14}=\frac{-14}{28} =-1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+17}{2*14}=\frac{20}{28} =5/7 $
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