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3t^2-14t+8=0
a = 3; b = -14; c = +8;
Δ = b2-4ac
Δ = -142-4·3·8
Δ = 100
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{100}=10$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-10}{2*3}=\frac{4}{6} =2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+10}{2*3}=\frac{24}{6} =4 $
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