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3t^2-6t-3=0
a = 3; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·3·(-3)
Δ = 72
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}
The end solution:
\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{2}}{2*3}=\frac{6-6\sqrt{2}}{6}t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{2}}{2*3}=\frac{6+6\sqrt{2}}{6}
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