6t^2+9t+3=0

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Solution for 6t^2+9t+3=0 equation: 



6t^2+9t+3=0

a = 6; b = 9; c = +3;
Δ = b2-4ac
Δ = 92-4·6·3
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*6}=\frac{-12}{12}
=-1
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*6}=\frac{-6}{12}
=-1/2
$

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