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1.6t^2=20t
We move all terms to the left:
1.6t^2-(20t)=0
a = 1.6; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·1.6·0
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*1.6}=\frac{0}{3.2} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*1.6}=\frac{40}{3.2} =12+1/2 $
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