2t^2+5t-13=0

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Solution for 2t^2+5t-13=0 equation: 



2t^2+5t-13=0

a = 2; b = 5; c = -13;
Δ = b2-4ac
Δ = 52-4·2·(-13)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{129}}{2*2}=\frac{-5-\sqrt{129}}{4}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{129}}{2*2}=\frac{-5+\sqrt{129}}{4}
$

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