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-3t^2-18t=0
a = -3; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·(-3)·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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*-3}=\frac{0}{-6} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*-3}=\frac{36}{-6} =-6 $
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