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8z^2-14z+5=0
a = 8; b = -14; c = +5;
Δ = b2-4ac
Δ = -142-4·8·5
Δ = 36
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{36}=6$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*8}=\frac{8}{16} =1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*8}=\frac{20}{16} =1+1/4 $
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