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28t^2-17t=0
a = 28; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·28·0
Δ = 289
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{289}=17$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*28}=\frac{0}{56} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*28}=\frac{34}{56} =17/28 $
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