2c^2+3c+1=0

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Solution for 2c^2+3c+1=0 equation:



2c^2+3c+1=0
a = 2; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*2}=\frac{-4}{4} =-1 $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*2}=\frac{-2}{4} =-1/2 $

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