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5t^2-96t+172=0
a = 5; b = -96; c = +172;
Δ = b2-4ac
Δ = -962-4·5·172
Δ = 5776
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{5776}=76$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-76}{2*5}=\frac{20}{10} =2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+76}{2*5}=\frac{172}{10} =17+1/5 $
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