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0.2t^2+3t=10.4
We move all terms to the left:
0.2t^2+3t-(10.4)=0
We add all the numbers together, and all the variables
0.2t^2+3t-10.4=0
a = 0.2; b = 3; c = -10.4;
Δ = b2-4ac
Δ = 32-4·0.2·(-10.4)
Δ = 17.32
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{-(3)-\sqrt{17.32}}{2*0.2}=\frac{-3-\sqrt{17.32}}{0.4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17.32}}{2*0.2}=\frac{-3+\sqrt{17.32}}{0.4} $
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