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2t^2-50t-50000=0
a = 2; b = -50; c = -50000;
Δ = b2-4ac
Δ = -502-4·2·(-50000)
Δ = 402500
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{402500}=\sqrt{2500*161}=\sqrt{2500}*\sqrt{161}=50\sqrt{161}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50\sqrt{161}}{2*2}=\frac{50-50\sqrt{161}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50\sqrt{161}}{2*2}=\frac{50+50\sqrt{161}}{4} $
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