2t^2+11t+9=0

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Solution for 2t^2+11t+9=0 equation: 



2t^2+11t+9=0

a = 2; b = 11; c = +9;
Δ = b2-4ac
Δ = 112-4·2·9
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*2}=\frac{-18}{4}
=-4+1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*2}=\frac{-4}{4}
=-1
$

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