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3t^2-243t=0
a = 3; b = -243; c = 0;
Δ = b2-4ac
Δ = -2432-4·3·0
Δ = 59049
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{59049}=243$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-243)-243}{2*3}=\frac{0}{6} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-243)+243}{2*3}=\frac{486}{6} =81 $
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