2t^2+1+3t=0

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



2t^2+1+3t=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:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


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


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

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