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 = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


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


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

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