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3t^2-19t+20=0
a = 3; b = -19; c = +20;
Δ = b2-4ac
Δ = -192-4·3·20
Δ = 121
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{121}=11$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*3}=\frac{8}{6} =1+1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*3}=\frac{30}{6} =5 $
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