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5z^2-14z+9=0
a = 5; b = -14; c = +9;
Δ = b2-4ac
Δ = -142-4·5·9
Δ = 16
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{16}=4$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*5}=\frac{10}{10} =1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*5}=\frac{18}{10} =1+4/5 $
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