2t2+5t-3=0

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Solution for 2t2+5t-3=0 equation: 



2t^2+5t-3=0

a = 2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·2·(-3)
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{-(5)-7}{2*2}=\frac{-12}{4}
=-3
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*2}=\frac{2}{4}
=1/2
$

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