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-3t^2+4t+3.9=0
a = -3; b = 4; c = +3.9;
Δ = b2-4ac
Δ = 42-4·(-3)·3.9
Δ = 62.8
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{-(4)-\sqrt{62.8}}{2*-3}=\frac{-4-\sqrt{62.8}}{-6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{62.8}}{2*-3}=\frac{-4+\sqrt{62.8}}{-6} $
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