18y^2+27y+10=0

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Solution for 18y^2+27y+10=0 equation: 



18y^2+27y+10=0

a = 18; b = 27; c = +10;
Δ = b2-4ac
Δ = 272-4·18·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3}{2*18}=\frac{-30}{36}
=-5/6
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3}{2*18}=\frac{-24}{36}
=-2/3
$

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