6t2+5t+1=0

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Solution for 6t2+5t+1=0 equation: 



6t^2+5t+1=0

a = 6; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·6·1
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{-(5)-1}{2*6}=\frac{-6}{12}
=-1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*6}=\frac{-4}{12}
=-1/3
$

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