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3t^2-8t+5=0
a = 3; b = -8; c = +5;
Δ = b2-4ac
Δ = -82-4·3·5
Δ = 4
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{4}=2$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*3}=\frac{6}{6} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*3}=\frac{10}{6} =1+2/3 $
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