9z^2-8+z=0

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Solution for 9z^2-8+z=0 equation: 



9z^2-8+z=0

a = 9; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·9·(-8)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*9}=\frac{-18}{18}
=-1
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*9}=\frac{16}{18}
=8/9
$

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