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14z^2+9z-5=0
a = 14; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·14·(-5)
Δ = 361
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{361}=19$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-19}{2*14}=\frac{-28}{28} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+19}{2*14}=\frac{10}{28} =5/14 $
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