2t2+t-3=0

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Solution for 2t2+t-3=0 equation: 



2t^2+t-3=0

a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4}
=-1+1/2
$


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

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