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3t^2-40t+100=0
a = 3; b = -40; c = +100;
Δ = b2-4ac
Δ = -402-4·3·100
Δ = 400
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{400}=20$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*3}=\frac{20}{6} =3+1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*3}=\frac{60}{6} =10 $
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