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4t-0.8t^2=0
a = -0.8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-0.8)·0
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-0.8}=\frac{-8}{-1.6} =+5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-0.8}=\frac{0}{-1.6} =0 $
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