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0.8t^2-4t+2.5=0
a = 0.8; b = -4; c = +2.5;
Δ = b2-4ac
Δ = -42-4·0.8·2.5
Δ = 8
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*0.8}=\frac{4-2\sqrt{2}}{1.6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*0.8}=\frac{4+2\sqrt{2}}{1.6} $
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