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t^2-40t-100=0
a = 1; b = -40; c = -100;
Δ = b2-4ac
Δ = -402-4·1·(-100)
Δ = 2000
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{5}}{2*1}=\frac{40-20\sqrt{5}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{5}}{2*1}=\frac{40+20\sqrt{5}}{2} $
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