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.2t^2-5t+17=0
a = .2; b = -5; c = +17;
Δ = b2-4ac
Δ = -52-4·.2·17
Δ = 11.4
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{-(-5)-\sqrt{11.4}}{2*.2}=\frac{5-\sqrt{11.4}}{0.4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{11.4}}{2*.2}=\frac{5+\sqrt{11.4}}{0.4} $
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