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10t^2-50t=0
a = 10; b = -50; c = 0;
Δ = b2-4ac
Δ = -502-4·10·0
Δ = 2500
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{2500}=50$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50}{2*10}=\frac{0}{20} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50}{2*10}=\frac{100}{20} =5 $
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