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5t^2-49t-320=0
a = 5; b = -49; c = -320;
Δ = b2-4ac
Δ = -492-4·5·(-320)
Δ = 8801
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{-(-49)-\sqrt{8801}}{2*5}=\frac{49-\sqrt{8801}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{8801}}{2*5}=\frac{49+\sqrt{8801}}{10} $
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