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5t^2-27t=0
a = 5; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·5·0
Δ = 729
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{729}=27$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27}{2*5}=\frac{0}{10} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*5}=\frac{54}{10} =5+2/5 $
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