18z^2+9z+1=0

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Solution for 18z^2+9z+1=0 equation: 



18z^2+9z+1=0

a = 18; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·18·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*18}=\frac{-12}{36}
=-1/3
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*18}=\frac{-6}{36}
=-1/6
$

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