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-12t^2+18t=0
a = -12; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-12)·0
Δ = 324
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{324}=18$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-12}=\frac{-36}{-24} =1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-12}=\frac{0}{-24} =0 $
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