2t2+9t+4=0

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Solution for 2t2+9t+4=0 equation: 



2t^2+9t+4=0

a = 2; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·2·4
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{-(9)-7}{2*2}=\frac{-16}{4}
=-4
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*2}=\frac{-2}{4}
=-1/2
$

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