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5z^2-6z-1=0
a = 5; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·5·(-1)
Δ = 56
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{14}}{2*5}=\frac{6-2\sqrt{14}}{10} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{14}}{2*5}=\frac{6+2\sqrt{14}}{10} $
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