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3t^2-3t-20=0
a = 3; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·3·(-20)
Δ = 249
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{-(-3)-\sqrt{249}}{2*3}=\frac{3-\sqrt{249}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{249}}{2*3}=\frac{3+\sqrt{249}}{6} $
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