3t2+t-3=0

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Solution for 3t2+t-3=0 equation: 



3t^2+t-3=0

a = 3; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·3·(-3)
Δ = 37
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{-(1)-\sqrt{37}}{2*3}=\frac{-1-\sqrt{37}}{6}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{37}}{2*3}=\frac{-1+\sqrt{37}}{6}
$

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