5z2+6z+1=0

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Solution for 5z2+6z+1=0 equation: 



5z^2+6z+1=0

a = 5; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·5·1
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{-(6)-4}{2*5}=\frac{-10}{10}
=-1
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*5}=\frac{-2}{10}
=-1/5
$

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