t^2+13t+19=0

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Solution for t^2+13t+19=0 equation: 



t^2+13t+19=0

a = 1; b = 13; c = +19;
Δ = b2-4ac
Δ = 132-4·1·19
Δ = 93
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{-(13)-\sqrt{93}}{2*1}=\frac{-13-\sqrt{93}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{93}}{2*1}=\frac{-13+\sqrt{93}}{2}
$

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