6(t^2)+11t+4=0

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Solution for 6(t^2)+11t+4=0 equation: 



6(t^2)+11t+4=0

a = 6; b = 11; c = +4;
Δ = b2-4ac
Δ = 112-4·6·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*6}=\frac{-16}{12}
=-1+1/3
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*6}=\frac{-6}{12}
=-1/2
$

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