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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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