2e^2+9e+3=0

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Solution for 2e^2+9e+3=0 equation: 



2e^2+9e+3=0

a = 2; b = 9; c = +3;
Δ = b2-4ac
Δ = 92-4·2·3
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{57}}{2*2}=\frac{-9-\sqrt{57}}{4}
$


$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{57}}{2*2}=\frac{-9+\sqrt{57}}{4}
$

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