3t2+19t+26=0

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Solution for 3t2+19t+26=0 equation:



3t^2+19t+26=0
a = 3; b = 19; c = +26;
Δ = b2-4ac
Δ = 192-4·3·26
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{-(19)-7}{2*3}=\frac{-26}{6} =-4+1/3 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+7}{2*3}=\frac{-12}{6} =-2 $

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