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1.8t^2-18t=0
a = 1.8; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·1.8·0
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*1.8}=\frac{0}{3.6} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*1.8}=\frac{36}{3.6} =10 $
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