5t^2-19t-4=0

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Solution for 5t^2-19t-4=0 equation:



5t^2-19t-4=0
a = 5; b = -19; c = -4;
Δ = b2-4ac
Δ = -192-4·5·(-4)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*5}=\frac{-2}{10} =-1/5 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*5}=\frac{40}{10} =4 $

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