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0.05t^2=2t+4.4
We move all terms to the left:
0.05t^2-(2t+4.4)=0
We get rid of parentheses
0.05t^2-2t-4.4=0
a = 0.05; b = -2; c = -4.4;
Δ = b2-4ac
Δ = -22-4·0.05·(-4.4)
Δ = 4.88
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-\sqrt{4.88}}{2*0.05}=\frac{2-\sqrt{4.88}}{0.1} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{4.88}}{2*0.05}=\frac{2+\sqrt{4.88}}{0.1} $
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