18p^2+33p+9=0

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Solution for 18p^2+33p+9=0 equation: 



18p^2+33p+9=0

a = 18; b = 33; c = +9;
Δ = b2-4ac
Δ = 332-4·18·9
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-21}{2*18}=\frac{-54}{36}
=-1+1/2
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+21}{2*18}=\frac{-12}{36}
=-1/3
$

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