(5t^2)-9t+3=0

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Solution for (5t^2)-9t+3=0 equation: 



(5t^2)-9t+3=0

a = 5; b = -9; c = +3;
Δ = b2-4ac
Δ = -92-4·5·3
Δ = 21
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{-(-9)-\sqrt{21}}{2*5}=\frac{9-\sqrt{21}}{10}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{21}}{2*5}=\frac{9+\sqrt{21}}{10}
$

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