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.5t^2-20t=0
a = .5; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·.5·0
Δ = 400
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{400}=20$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*.5}=\frac{0}{1} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*.5}=\frac{40}{1} =40 $
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